template - WordPress.com · Web viewFP1 PAST EXAM QUESTIONS ON INDUCTION WITH 1. Use the method of...

Algebra skills for proof by inductionPlease fill in the blanks then mark your answers against the answer sheet

1 ab + ac = a (___ + ___)

2 ab2 + a2b = ab (___ + ___)

3 ab + a3 = a (___ + ___)

4 frac14ab + ac = frac14a (___ + ___)

5 frac14ca2 = _____ times [frac14a2]

6 a3 = frac14a2 times _____

7 frac14k2a2 = _____ times [frac14a2]

8 frac14k2(k+1)2 = _____ times [frac14(k+1)2]

9 frac14k2a2 + a3 = [frac14a2] times (______ + ______)

10 frac14k2(k+1)2 + (k+1)3 = [frac14(k+1)2] times (________ + _________)

11 frac12abc + b2 = frac12b times (_______ + _______)

12 frac12n(n+1)(2n+1)+(n+1)2 = frac12(n+1)times(________ + ____________)

13 3a+b = ____ times ____

14 34k+4 = ____ times 34k

15 34(k+1) = ____ times 34k

16 24(k+1) = _____ times 24k

Workbook for proofs by induction

Name

17 71 times 7k = _______

18 7 times 7k = _______

19 7k+1 = ____ times 7k

20 5k+1 = ____ times 5k

21 ck+1 = ____ times ck

22 a times 7k+1 = ___________ times 7k

23 b times 7k+1 = ___________ times 7k

24 (2k+3) times 7k+1 = ___________ times 7k

25 7 times (2k+3) times 7k = 7 times (2k+1) times 7k + ________

26 Does 7 times 42 = 282

27 Does 9 times 42 = 362

28 Does 9 times 4k = 36k for any k except k=1


30 4k+1 = ____ times 4k

31 5k+1 = ____ times 5k

32 bk+1 = ____ times bk

33 7k+1 minus 4k+1 = ____ times 7k minus ____ times 4k



36 ak+1 minus bk+1 = ____ times ak minus ____ times bk

37 7a minus 4b = 7(a minus b) + ___ b minus 4b = 7(a minus b) + ___ b

38 7c minus 4d = 7(c minus d) + ___ d minus 4d = 7(c minus d) + ___ d

39 7 times 7k minus 4 times 4k = 7_____ minus 4______

40 9 times 9k minus 2 times 2k = 9____ minus 2_________





45 (k+1)3 = k3 + ___ k2 + ____ k + ___

46 (k+1)4 = k4 + ____ k3 + ______ k2 + ____ k + ___

47 (k+1)5 = k5 + _____ k4 + ______ k3 + ____ k2 + ____ k + ____

48 72k+1 = _____ times 72kminus1

Take out common factors and simplify these

49 16n(n+1)(2n+1) + n(n+1)

50 16n(n+1)(2n+1) + 2n(n+1)

51 frac14n2(n+1)2 + 16n(n+1)(2n+1) + n(n+1)

52 2frasl4n2(n+1)2 + 16n(n+1)(2n+1)

53 frac14n2(n+1)2 + 2frasl6n(n+1)(2n+1)

54 frac14n2(n+1)2 + 3frasl6n(n+1)(2n+1) + n(n+1)

55 16n(n+1)(2n+1) + n

56 16n(n+1)(2n+1) minus n

57 16n(n+1)(2n+1) minus 2n

58 16n(n+1)(2n+1) + frac12n(n+1) minus n

59 frac12(k+1)(k+2) minus frac12k(k+1)

60 16(k+1)(k+2)(2k+3) minus 16k(k+1)(2k+1)

61 frac14(k+1)2(k+2)2 minus frac14k2(k+1)2

Answers

1) ab + ac = a (b + c)

2) ab2 + a2b = ab (b + a)

3) ab + a3 = a (b + a2)

4) frac14ab + ac = frac14a (b + 4c)

5) frac14ca2 = c times [frac14a2]

6) a3 = frac14a2 times 4a

7) frac14k2a2 = k2 times [frac14a2]

8) frac14k2(k+1)2 = k2times [frac14(k+1)2]

9) frac14k2a2 + a3 = [frac14a2] times (k2 + 4a)

10) frac14k2(k+1)2+(k+1)3 = [frac14(k+1)2]times(k2+4[k+1])

11) frac12abc + b2 = frac12b times (ac + 2b)

12) frac12n(n+1)(2n+1)+(n+1)2 = frac12(n+1)times(n[2n+1] + 2[n+1])

13) 3a+b = 3a times 3b

14) 34k+4 =34 times 34k

15) 34(k+1) = 34 times 34k

16) 24(k+1) = 24 times 24k

17) 71 times 7k = 7k+1

18) 7 times 7k = 7k+1

19) 7k+1 = 7 times 7k

20) 5k+1 =5 times 5k

21) ck+1 = c times ck

22) a times 7k+1 = 7a times 7k

23) b times 7k+1 = 7b times 7k

24) (2k+3) times 7k+1 = 7(2k+3) times 7k

25) 7times(2k+3)times7k = 7times(2k+1)times7k + 7times2times7k

26) Does 7 times 42 = 282 No

27) Does 9 times 42 = 362 No

28) Does 9 times 4k = 36k for any k except k=1 No


30) 4k+1 = 4 times 4k

31) 5k+1 = 5 times 5k

32) bk+1 = b times bk

33) 7k+1 minus 4k+1 = 7 times 7k minus 4 times 4k



36) ak+1 minus bk+1 = a times ak minus b times bk

37) 7aminus4b = 7(aminusb)+7bminus4b = 7(aminusb) + 3b

38) 7cminus4d = 7(cminusd)+7dminus4d = 7(cminusd) + 3d

39) 7 times 7k minus 4 times 4k = 7k+1 minus 4k+1






45) (k+1)3 = k3 + 3 k2 + 3 k + 1

46) (k+1)4 = k4 + 4 k3 + 6 k2 + 4 k + 1

47) (k+1)5 = k5+5k4+10k3+10k2+5k+1

48) 72k+1 = 72 times 72kminus1

49) 16n(n+1)[2n+7]

50) 16n(n+1)[2n+13]

51) 1frasl12n(n+1)[3+2(2n+1) + 12] = 112n(n+1)[4n+17]

52) 16n(n+1)[3n(n+1)+(2n+1)] = 16n(n+1)[3n2+5n+1)]

53) 112n(n+1)[3n(n+1) + 4(2n+1)] = 112n(n+1)[3n2+11n+4)]

54) frac14n(n+1)[n(n+1) + 2(2n+1) + 4] = frac14n(n+1)[n2+5n+6]

55) 16n[(n+1)(2n+1) + 6] = 16n[2n2+3n+7]

56) 16n[(n+1)(2n+1) minus 6] = 16n[2n2+3nminus5]


58) 16n[(n+1)(2n+1) + 3(n+1) minus 6] = 16n[2n2+6nminus2] = 13n[n2+3nminus1]

59) k+1

60) 16(k+1)[(k+2)(2k+3) minus k(2k+1)] = 16(k+1)[6k+6] = (k+1)2

61) frac14(k+1)2[(k+2)2 minus k2] = (k+1)3

More practice if you need it

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20151023taking-out-common-factors

PROOF BY INDUCTION

1 A hotel has an infinite number of rooms numbered 1 2 3 The walls are thin so each guest can hear whats happening next door As soon as the guest in room n+1 hears the guest in room n going for dinner she or he gets hungry and goes for dinner too What will happen after the guest in room 1 goes for dinner

2 An infinitely large class of students are in a boring lesson in seats numbered 1 2 3 As soon as the student in seat n yawns the student in seat n+1 has an irresistible impulse to yawn What happens after the student in seat 1 yawns

3 An infinitely long queue of students is waiting to get lunch in the canteen A student in line will laugh if the student in front of her or him laughs If the first student in line laughs is there a student that will not eventually laugh

4 The same example but now a student in line will laugh if two or more students ahead of her or him laughs The first student in line laughs what happens now Can you be sure every student ends up laughing

5 Some positive integers are written on the board At

each step one of the integers n is erased and replaced with a positive integer less than n mdash unless the erased integer is 1 in which case it is not replaced Prove that no matter how this procedure is carried out the board must eventually become empty

6 Same thing but erased integers bigger than 1 are replaced with any number of positive integers all less than the erased one and theres a rule that you can never have the same integer repeated on the board Prove that the board must eventually become empty

7 Same things but drop the never repeat an integer rule Prove that the board must eventually become empty

Understanding the Σ signThis bit of maths

means the total you get by adding the contents of a series of trolleys

The Σ sign (pronounced sigma Greek capital S) tells you add up

The r=1 (or whatever) bit under the Σ sign tells you start adding with the 1st (or whatever) trolley

The n (or whatever) bit on top of the Σ sign tells you stop adding after the nrsquoth trolley

The bit after the Σ sign r2 in the example above tells you the formula for the contents of the rrsquoth trolley whatever r is

So if it is r2 after the Σ sign the 1st trolley has 1 unit on it the 2nd trolley has 22=4 units on it the 3rd trolley has 32=9 units on it the 4th trolley has 42=16 units on it

1 What is

2 What is

3 What is

4 An exam paper has 10 questions and you

get mk marks on question k What is

5 If you get 66 marks on the first nine questions and 7 marks on question 10

a What is

b What is

6 If you get 66 marks on the first nine

questions and p marks on question 10 what is the formula for your total marks

7 Fill in the blank

+______________

Answers1 1+4+9+16+25=552 1+8+27+64+125=2253 1+2+3=64 Your total marks for the whole paper5 a 66 b 736 66+p7 +m10

Scaffolded proofs by induction with series fill in the blanksℕ also called ℤ+ is the set of counting numbers 1 2 3 4 5

Claim = frac12n(n+1)

STEP 1 PROVE TRUE FOR n=1 1 = frac12 ⤫ 1⤫ 2

STEP 2 PROVE TRUE FOR n=k rArr TRUE FOR n=k+1

To prove = frac12 (k+1)(k+2)

True for n=k rArr = frac12 k(k+1)

rArr = frac12[k+2](k+1)

The claim is true for n = 1For all k we have shown that if it is true for n = k then it is true for n = k + 1So by mathematical induction it is true for all n isin1048605 ℕ

Claim = frac14n2(n+1)2



To prove

True for n=k rArr

So

= frac14(k+1)2[k2+4(k+1)] = frac14(k+1)2[k2+4k+4] = frac14(k+1)2(k+2)2

The claim is true for n = 1For all k we have shown that if it is true for n = k then it is true for n = k + 1

So by mathematical induction it is true for all n isin1048605 ℕ

Claim

Proof by induction STEP 1 PROVE TRUE FOR n=1


To prove __________________________________________

True for n=k rArr ______________________________________

so

= __________________________________ [take out the common factor]

=

The claim is true for n = 1 For all k we have shown that if it is true for n = k then it is true for n = k + 1 So by mathematical induction it is

true for all n isin1048605 ℕ

Claim



To prove __________________________________________

True for n=k rArr ______________________________________

+ ________________ [fill in last bit on this line]

so

= _______________________________________________________ [take out common factor]

= _______________________________________________________



Claim

Proof by induction

STEP 1 PROVE TRUE FOR n=1


To prove __________________________________________

True for n=k rArr ______________________________________

+ ___________________________

so


= _______________________________________________________



Claim



To prove __________________________________________

True for n=k rArr ______________________________________

+ _________________________________

so


= _______________________________________________________



FP1 PAST EXAM QUESTIONS ON INDUCTION WITH SERIES

1 Prove by induction that for n

n

r nn

rr1 1)1(1

(5)

2 Prove by induction that for n + 1212

3112

1

2

n

r

nnnr

(5)

3 Show that

n

r

r ndashr1

2 )1ndash ( = 3

1

(n ndash 2)n(n + 2)(6)

4 (a) Prove by induction that

12161

1

2

nnnrn

r

5 (a) Prove by induction that for any positive integer n

22

1

3 )1(41

nnrn

r

(5)

RULES ABOUT DIVISIBILITY

a | b means a divides exactly into b

So 2 | 6 3| 6 but 4 | 6 and 5 | 6

If we know for example that 31 | 3379 what else can we deduce

1 We can add (or subtract) any number of 31s to the right hand side so for example

31 | 3379 + 31 in other words 31 | 3410

and 31 | 3379 + 11 times 31 in other words 31 | 3720

2 We can multiply the right hand side by any number so eg

31 | 10 times 3379 in other words 31 | 33790

Tick or cross to show whether these statements are true or false

1 11 | 110 rArr 11 | 143 = 110 + 3 times 11

2 7 | 60 + 3 rArr 7 | (60 + 3) times 10 = 630

3 7 | 60 + 3 rArr 7 | 60 + 3times10 = 90

Write down some correct conclusions you can make from the following true statements

1 13 | 91

2 14 | 210

3 15 | 300

4 16 | 1024

5 17 | 306

6 109 | 3379

SCAFFOLDED PROOFS BY INDUCTION (DIVISIBILITY)

p | q means p divides exactly into q eg 2 | 6

[1] Claim for all n 6 | 7nminus1

STEP 1 Prove true for n = 1

STEP 2 Prove true for n=k rArr true for n=k+1To prove 6 | 7k+1minus1True for n=k rArr 6 | 7kminus1rArr 6| 7k+6times7kminus1 (+6times7k)rArr 6 | 7times7kminus1rArr 6 | 7k+1minus1emspemsp The claim is true for n = 1For all k we have shown that if it is true for n = k then it is true for n = k + 1So by mathematical induction it is true for all n isin 1048605ℕ

[2] Claim for all n 10 | 72nminus1 + 32nminus1


STEP 2 Prove true for n=k rArr true for n=k+1To prove 10 | 72k+1 + 32k+1

True for n=k rArr 10 | 72kminus1 + 32kminus1

rArr 10| 72times72kminus1 + 72times32kminus1 (times 72)rArr 10 | 72k+1+9times32kminus1+40times32kminus1 rArr 10 | 72k+1 + 32k+1emspemsp ( minus 10times4times32kminus1)The claim is true for n = 1For all k we have shown that if it is true for n = k then it is true for n = k + 1So by mathematical induction it is true for all n isin1048605 ℕ

[3] Claim for all n 3 | 7nminus 4n


STEP 2 Prove true for n=k rArr true for n=k+1To prove 3 | 7k+1minus 4k+1

True for n=k rArr 3 | 7kminus 4k

rArr 3| 7times7kminus7times4k (times 7)rArr 3 | 7k+1minus 4times4k minus __________rArr 3 | 7k+1minus 4times4k (+ 3times4k)rArr 3 | 7k+1minus 4k+1emsp ________________________ rArr _____________________________ The claim is true for n = 1For all k we have shown that if it is true for n = k then it is true for n = k + 1So by mathematical induction it is true for all n isin1048605 ℕ



STEP 2 Prove true for n=k+1 if true for n=kTo prove 5 | 9k+1minus 4k+1

True for n=k rArr 5 |________________rArr 5| 9times9kminus9times4k (times 9)rArr 5 | 9k+1 minus ____times4k minus ___times4k rArr 5 | 9k+1minus 4k+1emsp (+ 5 times4k) The claim is true for n = 1For all k we have shown that if it is true for n = k then it is true for n = k + 1So by mathematical induction it is true for all n isin1048605 ℕ





rArr 5| _____times7k minus_____times2k (times ___)rArr 5 | _____times7k minus_____times2k minus_____times2k

rArr 5 | 7k+1minus 2k+1emsp (+5 times2k)The claim is true for n = 1For all k we have shown that if it is true for n = k then it is true for n = k + 1So by mathematical induction it is true for all n isin1048605 ℕ

[5] Claim for all n 9 | 72n+1+ 22n+1


STEP 2 Prove true for n=k+1 if true for n=kTo prove 9 | 72(k+1)+1+ 22(k+1)+1 ie 9 | 72k+3+ 22k+3

True for n=k rArr 9 | 72k+1+ 22k+1

rArr 9 | 72times72k+1+72times22k+1 (times ____)rArr 9 | 72k+3 + ______times22k+1 + 45times22k+1

rArr 9 | 72k+3+ 22k+3emsp ( minus 9times5times22k+1)The claim is true for n = 1For all k we have shown that if it is true for n = k then it is true for n = k + 1So by mathematical induction it is true for all n isin1048605 ℕ

[7] Claim for all n 13 | 72n+1+ 62n+1



True for n=k rArr 13 | __________________rArr 13| ______ times72k+1 + ______times 62k+1 (times _______)rArr 13 | 72k+3 + ______times 62k+1 + ______times 62k+1

rArr 13 | 72k+3+62k+3emsp ( minus ________)


[8] Claim for all n 5 | n5minusn STEP 1 Prove true for n = 1 STEP 2 Prove true for n=k+1 if true for n=kTo prove 5 | (k+1)5minus(k+1)ie 5 | k5+5k4+10k3+10k2+5k+1minus(k+1)True for n=k rArr 5 | k5minuskrArr 5| k5minusk+ 5k4+10k3+10k2+5k (+5times _______)rArr 5| k5+ 5k4+10k3+10k2+5k +1minus(k+1) rArr 5| ______________________________ The claim is true for n = 1For all k we have shown that if it is true for n = k then it is true for n = k + 1So by mathematical induction it is true for all n isin1048605 ℕ

[9] Claim for all n 3 | n3minusn


STEP 2 Prove true for n=k+1 if true for n=kTo prove 3| ____________________________ie 3| k3+_____k2+________k + 1 minus (k+1)

True for n=k rArr 3| _____________

rArr 3| k3+ _________________________ minusk (+3k2+3k)rArr 3| (k+1)3minus(k+1) The claim is true for n = 1For all k we have shown that if it is true for n = k then it is true for n = k + 1So by mathematical induction it is true for all n isin1048605 ℕ



STEP 2 Prove true for n=k rArr true for n=k+1To prove __________________True for n=k rArr ______________________

rArr 7| 7times_______ + ____________________ (+7times8k)rArr 7 | ______times8kminus1rArr 7 | 8k+1minus1emspemsp The claim is true for n = 1For all k we have shown that if it is true for n = k then it is true for n = k + 1So by mathematical induction it is true for all n isin1048605 ℕ

[11] Claim for all n 7 | 8n + 14n + 6


STEP 2 Prove true for n=k rArr true for n=k+1

To prove __________________

True for n=k rArr ______________________

rArr 7| 7times_______ + ____________________ (+7times8k + 14)

rArr 7 | ________________________________

rArr 7 | _____________________________emsp The claim is true for n = 1For all k we have shown that if it is true for n = k then it is true for n = k + 1So by mathematical induction it is true for all n isin1048605 ℕ

[12] Claim for all n 6 | 7n + 12n + 5


71+12times1 + 5 = 24 and 6|24

STEP 2 Prove true for n=k rArr true for n=k+1To prove 6 | 7k+1 + 12(k+1) + 5________________True for n=k rArr 6 | 7k + 12k + 5______________________rArr 6 | 6times7k + 7k + 12k + 5 (+6times7k)rArr 6 | 7times7k + 12k + 5 rArr 6 | 7k+1 + 12k + 5 rArr 6 | 7k+1 + 12k + 12 + 5 (+12)rArr 6 | 7k+1 + 12(k+1) + 5 The claim is true for n = 1For all k we have shown that if it is true for n = k then it is true for n = k + 1So by mathematical induction it is true for all n isin1048605 ℕ

FP1 PAST EXAM QUESTIONS ON INDUCTION WITH DIVISIBILITY

1 Use the method of mathematical induction to prove that for n isin +

(b) f(n) = 4n + 6n ndash 1 is divisible by 3(7)

2

f(n) = (2n + 1)7n ndash 1

Prove by induction that for all positive integers n f(n) is divisible by 4(Total 6 marks)

3 Given that

f(n) = 34n + 24n+2

(a) show that for k + f(k + 1) ndash f(k) is divisible by 15(4)

(b) prove that for n + f(n) is divisible by 5(3)

(c) show that it is not true that for all positive integers n f(n) is divisible by 15(1)

(Total 8 marks)

12 Prove by induction that for n isin

(a) f(n) = 5n + 8n + 3 is divisible by 4(7)

City of London Academy 30

17 71 times 7k = _______

18 7 times 7k = _______

19 7k+1 = ____ times 7k

20 5k+1 = ____ times 5k

21 ck+1 = ____ times ck

22 a times 7k+1 = ___________ times 7k

23 b times 7k+1 = ___________ times 7k

24 (2k+3) times 7k+1 = ___________ times 7k

25 7 times (2k+3) times 7k = 7 times (2k+1) times 7k + ________

26 Does 7 times 42 = 282

27 Does 9 times 42 = 362



30 4k+1 = ____ times 4k

31 5k+1 = ____ times 5k

32 bk+1 = ____ times bk













45 (k+1)3 = k3 + ___ k2 + ____ k + ___

46 (k+1)4 = k4 + ____ k3 + ______ k2 + ____ k + ___

47 (k+1)5 = k5 + _____ k4 + ______ k3 + ____ k2 + ____ k + ____

48 72k+1 = _____ times 72kminus1


49 16n(n+1)(2n+1) + n(n+1)

50 16n(n+1)(2n+1) + 2n(n+1)

51 frac14n2(n+1)2 + 16n(n+1)(2n+1) + n(n+1)

52 2frasl4n2(n+1)2 + 16n(n+1)(2n+1)

53 frac14n2(n+1)2 + 2frasl6n(n+1)(2n+1)


55 16n(n+1)(2n+1) + n

56 16n(n+1)(2n+1) minus n

57 16n(n+1)(2n+1) minus 2n



60 16(k+1)(k+2)(2k+3) minus 16k(k+1)(2k+1)


Answers

1) ab + ac = a (b + c)

2) ab2 + a2b = ab (b + a)

3) ab + a3 = a (b + a2)











14) 34k+4 =34 times 34k

15) 34(k+1) = 34 times 34k

16) 24(k+1) = 24 times 24k

17) 71 times 7k = 7k+1

18) 7 times 7k = 7k+1

19) 7k+1 = 7 times 7k

20) 5k+1 =5 times 5k




24) (2k+3) times 7k+1 = 7(2k+3) times 7k


26) Does 7 times 42 = 282 No

27) Does 9 times 42 = 362 No



30) 4k+1 = 4 times 4k

31) 5k+1 = 5 times 5k














45) (k+1)3 = k3 + 3 k2 + 3 k + 1

46) (k+1)4 = k4 + 4 k3 + 6 k2 + 4 k + 1

47) (k+1)5 = k5+5k4+10k3+10k2+5k+1


49) 16n(n+1)[2n+7]

50) 16n(n+1)[2n+13]

51) 1frasl12n(n+1)[3+2(2n+1) + 12] = 112n(n+1)[4n+17]

52) 16n(n+1)[3n(n+1)+(2n+1)] = 16n(n+1)[3n2+5n+1)]

53) 112n(n+1)[3n(n+1) + 4(2n+1)] = 112n(n+1)[3n2+11n+4)]


55) 16n[(n+1)(2n+1) + 6] = 16n[2n2+3n+7]




59) k+1

60) 16(k+1)[(k+2)(2k+3) minus k(2k+1)] = 16(k+1)[6k+6] = (k+1)2

61) frac14(k+1)2[(k+2)2 minus k2] = (k+1)3




PROOF BY INDUCTION
















1 What is

2 What is

3 What is




a What is

b What is



7 Fill in the blank

+______________













To prove

True for n=k rArr

So




Claim



To prove __________________________________________

True for n=k rArr ______________________________________

so


=



Claim



To prove __________________________________________

True for n=k rArr ______________________________________


so


= _______________________________________________________



Claim

Proof by induction



To prove __________________________________________

True for n=k rArr ______________________________________

+ ___________________________

so


= _______________________________________________________



Claim



To prove __________________________________________

True for n=k rArr ______________________________________

+ _________________________________

so


= _______________________________________________________





n

r nn

rr1 1)1(1

(5)


3112

1

2

n

r

nnnr

(5)

3 Show that

n

r

r ndashr1

2 )1ndash ( = 3

1

(n ndash 2)n(n + 2)(6)


12161

1

2

nnnrn

r


22

1

3 )1(41

nnrn

r

(5)



So 2 | 6 3| 6 but 4 | 6 and 5 | 6



31 | 3379 + 31 in other words 31 | 3410





1 11 | 110 rArr 11 | 143 = 110 + 3 times 11

2 7 | 60 + 3 rArr 7 | (60 + 3) times 10 = 630

3 7 | 60 + 3 rArr 7 | 60 + 3times10 = 90


1 13 | 91

2 14 | 210

3 15 | 300

4 16 | 1024

5 17 | 306

6 109 | 3379


























[5] Claim for all n 9 | 72n+1+ 22n+1






[7] Claim for all n 13 | 72n+1+ 62n+1










True for n=k rArr 3| _____________






[11] Claim for all n 7 | 8n + 14n + 6



To prove __________________

True for n=k rArr ______________________


rArr 7 | ________________________________


[12] Claim for all n 6 | 7n + 12n + 5


71+12times1 + 5 = 24 and 6|24





2

f(n) = (2n + 1)7n ndash 1


3 Given that

f(n) = 34n + 24n+2




(Total 8 marks)














45 (k+1)3 = k3 + ___ k2 + ____ k + ___

46 (k+1)4 = k4 + ____ k3 + ______ k2 + ____ k + ___

47 (k+1)5 = k5 + _____ k4 + ______ k3 + ____ k2 + ____ k + ____

48 72k+1 = _____ times 72kminus1


49 16n(n+1)(2n+1) + n(n+1)

50 16n(n+1)(2n+1) + 2n(n+1)

51 frac14n2(n+1)2 + 16n(n+1)(2n+1) + n(n+1)

52 2frasl4n2(n+1)2 + 16n(n+1)(2n+1)

53 frac14n2(n+1)2 + 2frasl6n(n+1)(2n+1)


55 16n(n+1)(2n+1) + n

56 16n(n+1)(2n+1) minus n

57 16n(n+1)(2n+1) minus 2n



60 16(k+1)(k+2)(2k+3) minus 16k(k+1)(2k+1)


Answers

1) ab + ac = a (b + c)

2) ab2 + a2b = ab (b + a)

3) ab + a3 = a (b + a2)











14) 34k+4 =34 times 34k

15) 34(k+1) = 34 times 34k

16) 24(k+1) = 24 times 24k

17) 71 times 7k = 7k+1

18) 7 times 7k = 7k+1

19) 7k+1 = 7 times 7k

20) 5k+1 =5 times 5k




24) (2k+3) times 7k+1 = 7(2k+3) times 7k


26) Does 7 times 42 = 282 No

27) Does 9 times 42 = 362 No



30) 4k+1 = 4 times 4k

31) 5k+1 = 5 times 5k














45) (k+1)3 = k3 + 3 k2 + 3 k + 1

46) (k+1)4 = k4 + 4 k3 + 6 k2 + 4 k + 1

47) (k+1)5 = k5+5k4+10k3+10k2+5k+1


49) 16n(n+1)[2n+7]

50) 16n(n+1)[2n+13]

51) 1frasl12n(n+1)[3+2(2n+1) + 12] = 112n(n+1)[4n+17]

52) 16n(n+1)[3n(n+1)+(2n+1)] = 16n(n+1)[3n2+5n+1)]

53) 112n(n+1)[3n(n+1) + 4(2n+1)] = 112n(n+1)[3n2+11n+4)]


55) 16n[(n+1)(2n+1) + 6] = 16n[2n2+3n+7]




59) k+1

60) 16(k+1)[(k+2)(2k+3) minus k(2k+1)] = 16(k+1)[6k+6] = (k+1)2

61) frac14(k+1)2[(k+2)2 minus k2] = (k+1)3




PROOF BY INDUCTION
















1 What is

2 What is

3 What is




a What is

b What is



7 Fill in the blank

+______________













To prove

True for n=k rArr

So




Claim



To prove __________________________________________

True for n=k rArr ______________________________________

so


=



Claim



To prove __________________________________________

True for n=k rArr ______________________________________


so


= _______________________________________________________



Claim

Proof by induction



To prove __________________________________________

True for n=k rArr ______________________________________

+ ___________________________

so


= _______________________________________________________



Claim



To prove __________________________________________

True for n=k rArr ______________________________________

+ _________________________________

so


= _______________________________________________________





n

r nn

rr1 1)1(1

(5)


3112

1

2

n

r

nnnr

(5)

3 Show that

n

r

r ndashr1

2 )1ndash ( = 3

1

(n ndash 2)n(n + 2)(6)


12161

1

2

nnnrn

r


22

1

3 )1(41

nnrn

r

(5)



So 2 | 6 3| 6 but 4 | 6 and 5 | 6



31 | 3379 + 31 in other words 31 | 3410





1 11 | 110 rArr 11 | 143 = 110 + 3 times 11

2 7 | 60 + 3 rArr 7 | (60 + 3) times 10 = 630

3 7 | 60 + 3 rArr 7 | 60 + 3times10 = 90


1 13 | 91

2 14 | 210

3 15 | 300

4 16 | 1024

5 17 | 306

6 109 | 3379


























[5] Claim for all n 9 | 72n+1+ 22n+1






[7] Claim for all n 13 | 72n+1+ 62n+1










True for n=k rArr 3| _____________






[11] Claim for all n 7 | 8n + 14n + 6



To prove __________________

True for n=k rArr ______________________


rArr 7 | ________________________________


[12] Claim for all n 6 | 7n + 12n + 5


71+12times1 + 5 = 24 and 6|24





2

f(n) = (2n + 1)7n ndash 1


3 Given that

f(n) = 34n + 24n+2




(Total 8 marks)




51 frac14n2(n+1)2 + 16n(n+1)(2n+1) + n(n+1)

52 2frasl4n2(n+1)2 + 16n(n+1)(2n+1)

53 frac14n2(n+1)2 + 2frasl6n(n+1)(2n+1)


55 16n(n+1)(2n+1) + n

56 16n(n+1)(2n+1) minus n

57 16n(n+1)(2n+1) minus 2n



60 16(k+1)(k+2)(2k+3) minus 16k(k+1)(2k+1)


Answers

1) ab + ac = a (b + c)

2) ab2 + a2b = ab (b + a)

3) ab + a3 = a (b + a2)











14) 34k+4 =34 times 34k

15) 34(k+1) = 34 times 34k

16) 24(k+1) = 24 times 24k

17) 71 times 7k = 7k+1

18) 7 times 7k = 7k+1

19) 7k+1 = 7 times 7k

20) 5k+1 =5 times 5k




24) (2k+3) times 7k+1 = 7(2k+3) times 7k


26) Does 7 times 42 = 282 No

27) Does 9 times 42 = 362 No



30) 4k+1 = 4 times 4k

31) 5k+1 = 5 times 5k














45) (k+1)3 = k3 + 3 k2 + 3 k + 1

46) (k+1)4 = k4 + 4 k3 + 6 k2 + 4 k + 1

47) (k+1)5 = k5+5k4+10k3+10k2+5k+1


49) 16n(n+1)[2n+7]

50) 16n(n+1)[2n+13]

51) 1frasl12n(n+1)[3+2(2n+1) + 12] = 112n(n+1)[4n+17]

52) 16n(n+1)[3n(n+1)+(2n+1)] = 16n(n+1)[3n2+5n+1)]

53) 112n(n+1)[3n(n+1) + 4(2n+1)] = 112n(n+1)[3n2+11n+4)]


55) 16n[(n+1)(2n+1) + 6] = 16n[2n2+3n+7]




59) k+1

60) 16(k+1)[(k+2)(2k+3) minus k(2k+1)] = 16(k+1)[6k+6] = (k+1)2

61) frac14(k+1)2[(k+2)2 minus k2] = (k+1)3




PROOF BY INDUCTION
















1 What is

2 What is

3 What is




a What is

b What is



7 Fill in the blank

+______________













To prove

True for n=k rArr

So




Claim



To prove __________________________________________

True for n=k rArr ______________________________________

so


=



Claim



To prove __________________________________________

True for n=k rArr ______________________________________


so


= _______________________________________________________



Claim

Proof by induction



To prove __________________________________________

True for n=k rArr ______________________________________

+ ___________________________

so


= _______________________________________________________



Claim



To prove __________________________________________

True for n=k rArr ______________________________________

+ _________________________________

so


= _______________________________________________________





n

r nn

rr1 1)1(1

(5)


3112

1

2

n

r

nnnr

(5)

3 Show that

n

r

r ndashr1

2 )1ndash ( = 3

1

(n ndash 2)n(n + 2)(6)


12161

1

2

nnnrn

r


22

1

3 )1(41

nnrn

r

(5)



So 2 | 6 3| 6 but 4 | 6 and 5 | 6



31 | 3379 + 31 in other words 31 | 3410





1 11 | 110 rArr 11 | 143 = 110 + 3 times 11

2 7 | 60 + 3 rArr 7 | (60 + 3) times 10 = 630

3 7 | 60 + 3 rArr 7 | 60 + 3times10 = 90


1 13 | 91

2 14 | 210

3 15 | 300

4 16 | 1024

5 17 | 306

6 109 | 3379


























[5] Claim for all n 9 | 72n+1+ 22n+1






[7] Claim for all n 13 | 72n+1+ 62n+1










True for n=k rArr 3| _____________






[11] Claim for all n 7 | 8n + 14n + 6



To prove __________________

True for n=k rArr ______________________


rArr 7 | ________________________________


[12] Claim for all n 6 | 7n + 12n + 5


71+12times1 + 5 = 24 and 6|24





2

f(n) = (2n + 1)7n ndash 1


3 Given that

f(n) = 34n + 24n+2




(Total 8 marks)




Answers

1) ab + ac = a (b + c)

2) ab2 + a2b = ab (b + a)

3) ab + a3 = a (b + a2)











14) 34k+4 =34 times 34k

15) 34(k+1) = 34 times 34k

16) 24(k+1) = 24 times 24k

17) 71 times 7k = 7k+1

18) 7 times 7k = 7k+1

19) 7k+1 = 7 times 7k

20) 5k+1 =5 times 5k




24) (2k+3) times 7k+1 = 7(2k+3) times 7k


26) Does 7 times 42 = 282 No

27) Does 9 times 42 = 362 No



30) 4k+1 = 4 times 4k

31) 5k+1 = 5 times 5k














45) (k+1)3 = k3 + 3 k2 + 3 k + 1

46) (k+1)4 = k4 + 4 k3 + 6 k2 + 4 k + 1

47) (k+1)5 = k5+5k4+10k3+10k2+5k+1


49) 16n(n+1)[2n+7]

50) 16n(n+1)[2n+13]

51) 1frasl12n(n+1)[3+2(2n+1) + 12] = 112n(n+1)[4n+17]

52) 16n(n+1)[3n(n+1)+(2n+1)] = 16n(n+1)[3n2+5n+1)]

53) 112n(n+1)[3n(n+1) + 4(2n+1)] = 112n(n+1)[3n2+11n+4)]


55) 16n[(n+1)(2n+1) + 6] = 16n[2n2+3n+7]




59) k+1

60) 16(k+1)[(k+2)(2k+3) minus k(2k+1)] = 16(k+1)[6k+6] = (k+1)2

61) frac14(k+1)2[(k+2)2 minus k2] = (k+1)3




PROOF BY INDUCTION
















1 What is

2 What is

3 What is




a What is

b What is



7 Fill in the blank

+______________













To prove

True for n=k rArr

So




Claim



To prove __________________________________________

True for n=k rArr ______________________________________

so


=



Claim



To prove __________________________________________

True for n=k rArr ______________________________________


so


= _______________________________________________________



Claim

Proof by induction



To prove __________________________________________

True for n=k rArr ______________________________________

+ ___________________________

so


= _______________________________________________________



Claim



To prove __________________________________________

True for n=k rArr ______________________________________

+ _________________________________

so


= _______________________________________________________





n

r nn

rr1 1)1(1

(5)


3112

1

2

n

r

nnnr

(5)

3 Show that

n

r

r ndashr1

2 )1ndash ( = 3

1

(n ndash 2)n(n + 2)(6)


12161

1

2

nnnrn

r


22

1

3 )1(41

nnrn

r

(5)



So 2 | 6 3| 6 but 4 | 6 and 5 | 6



31 | 3379 + 31 in other words 31 | 3410





1 11 | 110 rArr 11 | 143 = 110 + 3 times 11

2 7 | 60 + 3 rArr 7 | (60 + 3) times 10 = 630

3 7 | 60 + 3 rArr 7 | 60 + 3times10 = 90


1 13 | 91

2 14 | 210

3 15 | 300

4 16 | 1024

5 17 | 306

6 109 | 3379


























[5] Claim for all n 9 | 72n+1+ 22n+1






[7] Claim for all n 13 | 72n+1+ 62n+1










True for n=k rArr 3| _____________






[11] Claim for all n 7 | 8n + 14n + 6



To prove __________________

True for n=k rArr ______________________


rArr 7 | ________________________________


[12] Claim for all n 6 | 7n + 12n + 5


71+12times1 + 5 = 24 and 6|24





2

f(n) = (2n + 1)7n ndash 1


3 Given that

f(n) = 34n + 24n+2




(Total 8 marks)








45) (k+1)3 = k3 + 3 k2 + 3 k + 1

46) (k+1)4 = k4 + 4 k3 + 6 k2 + 4 k + 1

47) (k+1)5 = k5+5k4+10k3+10k2+5k+1


49) 16n(n+1)[2n+7]

50) 16n(n+1)[2n+13]

51) 1frasl12n(n+1)[3+2(2n+1) + 12] = 112n(n+1)[4n+17]

52) 16n(n+1)[3n(n+1)+(2n+1)] = 16n(n+1)[3n2+5n+1)]

53) 112n(n+1)[3n(n+1) + 4(2n+1)] = 112n(n+1)[3n2+11n+4)]


55) 16n[(n+1)(2n+1) + 6] = 16n[2n2+3n+7]




59) k+1

60) 16(k+1)[(k+2)(2k+3) minus k(2k+1)] = 16(k+1)[6k+6] = (k+1)2

61) frac14(k+1)2[(k+2)2 minus k2] = (k+1)3




PROOF BY INDUCTION
















1 What is

2 What is

3 What is




a What is

b What is



7 Fill in the blank

+______________













To prove

True for n=k rArr

So




Claim



To prove __________________________________________

True for n=k rArr ______________________________________

so


=



Claim



To prove __________________________________________

True for n=k rArr ______________________________________


so


= _______________________________________________________



Claim

Proof by induction



To prove __________________________________________

True for n=k rArr ______________________________________

+ ___________________________

so


= _______________________________________________________



Claim



To prove __________________________________________

True for n=k rArr ______________________________________

+ _________________________________

so


= _______________________________________________________





n

r nn

rr1 1)1(1

(5)


3112

1

2

n

r

nnnr

(5)

3 Show that

n

r

r ndashr1

2 )1ndash ( = 3

1

(n ndash 2)n(n + 2)(6)


12161

1

2

nnnrn

r


22

1

3 )1(41

nnrn

r

(5)



So 2 | 6 3| 6 but 4 | 6 and 5 | 6



31 | 3379 + 31 in other words 31 | 3410





1 11 | 110 rArr 11 | 143 = 110 + 3 times 11

2 7 | 60 + 3 rArr 7 | (60 + 3) times 10 = 630

3 7 | 60 + 3 rArr 7 | 60 + 3times10 = 90


1 13 | 91

2 14 | 210

3 15 | 300

4 16 | 1024

5 17 | 306

6 109 | 3379


























[5] Claim for all n 9 | 72n+1+ 22n+1






[7] Claim for all n 13 | 72n+1+ 62n+1










True for n=k rArr 3| _____________






[11] Claim for all n 7 | 8n + 14n + 6



To prove __________________

True for n=k rArr ______________________


rArr 7 | ________________________________


[12] Claim for all n 6 | 7n + 12n + 5


71+12times1 + 5 = 24 and 6|24





2

f(n) = (2n + 1)7n ndash 1


3 Given that

f(n) = 34n + 24n+2




(Total 8 marks)




PROOF BY INDUCTION
















1 What is

2 What is

3 What is




a What is

b What is



7 Fill in the blank

+______________













To prove

True for n=k rArr

So




Claim



To prove __________________________________________

True for n=k rArr ______________________________________

so


=



Claim



To prove __________________________________________

True for n=k rArr ______________________________________


so


= _______________________________________________________



Claim

Proof by induction



To prove __________________________________________

True for n=k rArr ______________________________________

+ ___________________________

so


= _______________________________________________________



Claim



To prove __________________________________________

True for n=k rArr ______________________________________

+ _________________________________

so


= _______________________________________________________





n

r nn

rr1 1)1(1

(5)


3112

1

2

n

r

nnnr

(5)

3 Show that

n

r

r ndashr1

2 )1ndash ( = 3

1

(n ndash 2)n(n + 2)(6)


12161

1

2

nnnrn

r


22

1

3 )1(41

nnrn

r

(5)



So 2 | 6 3| 6 but 4 | 6 and 5 | 6



31 | 3379 + 31 in other words 31 | 3410





1 11 | 110 rArr 11 | 143 = 110 + 3 times 11

2 7 | 60 + 3 rArr 7 | (60 + 3) times 10 = 630

3 7 | 60 + 3 rArr 7 | 60 + 3times10 = 90


1 13 | 91

2 14 | 210

3 15 | 300

4 16 | 1024

5 17 | 306

6 109 | 3379


























[5] Claim for all n 9 | 72n+1+ 22n+1






[7] Claim for all n 13 | 72n+1+ 62n+1










True for n=k rArr 3| _____________






[11] Claim for all n 7 | 8n + 14n + 6



To prove __________________

True for n=k rArr ______________________


rArr 7 | ________________________________


[12] Claim for all n 6 | 7n + 12n + 5


71+12times1 + 5 = 24 and 6|24





2

f(n) = (2n + 1)7n ndash 1


3 Given that

f(n) = 34n + 24n+2




(Total 8 marks)











1 What is

2 What is

3 What is




a What is

b What is



7 Fill in the blank

+______________













To prove

True for n=k rArr

So




Claim



To prove __________________________________________

True for n=k rArr ______________________________________

so


=



Claim



To prove __________________________________________

True for n=k rArr ______________________________________


so


= _______________________________________________________



Claim

Proof by induction



To prove __________________________________________

True for n=k rArr ______________________________________

+ ___________________________

so


= _______________________________________________________



Claim



To prove __________________________________________

True for n=k rArr ______________________________________

+ _________________________________

so


= _______________________________________________________





n

r nn

rr1 1)1(1

(5)


3112

1

2

n

r

nnnr

(5)

3 Show that

n

r

r ndashr1

2 )1ndash ( = 3

1

(n ndash 2)n(n + 2)(6)


12161

1

2

nnnrn

r


22

1

3 )1(41

nnrn

r

(5)



So 2 | 6 3| 6 but 4 | 6 and 5 | 6



31 | 3379 + 31 in other words 31 | 3410





1 11 | 110 rArr 11 | 143 = 110 + 3 times 11

2 7 | 60 + 3 rArr 7 | (60 + 3) times 10 = 630

3 7 | 60 + 3 rArr 7 | 60 + 3times10 = 90


1 13 | 91

2 14 | 210

3 15 | 300

4 16 | 1024

5 17 | 306

6 109 | 3379


























[5] Claim for all n 9 | 72n+1+ 22n+1






[7] Claim for all n 13 | 72n+1+ 62n+1










True for n=k rArr 3| _____________






[11] Claim for all n 7 | 8n + 14n + 6



To prove __________________

True for n=k rArr ______________________


rArr 7 | ________________________________


[12] Claim for all n 6 | 7n + 12n + 5


71+12times1 + 5 = 24 and 6|24





2

f(n) = (2n + 1)7n ndash 1


3 Given that

f(n) = 34n + 24n+2




(Total 8 marks)












To prove

True for n=k rArr

So




Claim



To prove __________________________________________

True for n=k rArr ______________________________________

so


=



Claim



To prove __________________________________________

True for n=k rArr ______________________________________


so


= _______________________________________________________



Claim

Proof by induction



To prove __________________________________________

True for n=k rArr ______________________________________

+ ___________________________

so


= _______________________________________________________



Claim



To prove __________________________________________

True for n=k rArr ______________________________________

+ _________________________________

so


= _______________________________________________________





n

r nn

rr1 1)1(1

(5)


3112

1

2

n

r

nnnr

(5)

3 Show that

n

r

r ndashr1

2 )1ndash ( = 3

1

(n ndash 2)n(n + 2)(6)


12161

1

2

nnnrn

r


22

1

3 )1(41

nnrn

r

(5)



So 2 | 6 3| 6 but 4 | 6 and 5 | 6



31 | 3379 + 31 in other words 31 | 3410





1 11 | 110 rArr 11 | 143 = 110 + 3 times 11

2 7 | 60 + 3 rArr 7 | (60 + 3) times 10 = 630

3 7 | 60 + 3 rArr 7 | 60 + 3times10 = 90


1 13 | 91

2 14 | 210

3 15 | 300

4 16 | 1024

5 17 | 306

6 109 | 3379


























[5] Claim for all n 9 | 72n+1+ 22n+1






[7] Claim for all n 13 | 72n+1+ 62n+1










True for n=k rArr 3| _____________






[11] Claim for all n 7 | 8n + 14n + 6



To prove __________________

True for n=k rArr ______________________


rArr 7 | ________________________________


[12] Claim for all n 6 | 7n + 12n + 5


71+12times1 + 5 = 24 and 6|24





2

f(n) = (2n + 1)7n ndash 1


3 Given that

f(n) = 34n + 24n+2




(Total 8 marks)




Claim



To prove __________________________________________

True for n=k rArr ______________________________________

so


=



Claim



To prove __________________________________________

True for n=k rArr ______________________________________


so


= _______________________________________________________



Claim

Proof by induction



To prove __________________________________________

True for n=k rArr ______________________________________

+ ___________________________

so


= _______________________________________________________



Claim



To prove __________________________________________

True for n=k rArr ______________________________________

+ _________________________________

so


= _______________________________________________________





n

r nn

rr1 1)1(1

(5)


3112

1

2

n

r

nnnr

(5)

3 Show that

n

r

r ndashr1

2 )1ndash ( = 3

1

(n ndash 2)n(n + 2)(6)


12161

1

2

nnnrn

r


22

1

3 )1(41

nnrn

r

(5)



So 2 | 6 3| 6 but 4 | 6 and 5 | 6



31 | 3379 + 31 in other words 31 | 3410





1 11 | 110 rArr 11 | 143 = 110 + 3 times 11

2 7 | 60 + 3 rArr 7 | (60 + 3) times 10 = 630

3 7 | 60 + 3 rArr 7 | 60 + 3times10 = 90


1 13 | 91

2 14 | 210

3 15 | 300

4 16 | 1024

5 17 | 306

6 109 | 3379


























[5] Claim for all n 9 | 72n+1+ 22n+1






[7] Claim for all n 13 | 72n+1+ 62n+1










True for n=k rArr 3| _____________






[11] Claim for all n 7 | 8n + 14n + 6



To prove __________________

True for n=k rArr ______________________


rArr 7 | ________________________________


[12] Claim for all n 6 | 7n + 12n + 5


71+12times1 + 5 = 24 and 6|24





2

f(n) = (2n + 1)7n ndash 1


3 Given that

f(n) = 34n + 24n+2




(Total 8 marks)




Claim



To prove __________________________________________

True for n=k rArr ______________________________________


so


= _______________________________________________________



Claim

Proof by induction



To prove __________________________________________

True for n=k rArr ______________________________________

+ ___________________________

so


= _______________________________________________________



Claim



To prove __________________________________________

True for n=k rArr ______________________________________

+ _________________________________

so


= _______________________________________________________





n

r nn

rr1 1)1(1

(5)


3112

1

2

n

r

nnnr

(5)

3 Show that

n

r

r ndashr1

2 )1ndash ( = 3

1

(n ndash 2)n(n + 2)(6)


12161

1

2

nnnrn

r


22

1

3 )1(41

nnrn

r

(5)



So 2 | 6 3| 6 but 4 | 6 and 5 | 6



31 | 3379 + 31 in other words 31 | 3410





1 11 | 110 rArr 11 | 143 = 110 + 3 times 11

2 7 | 60 + 3 rArr 7 | (60 + 3) times 10 = 630

3 7 | 60 + 3 rArr 7 | 60 + 3times10 = 90


1 13 | 91

2 14 | 210

3 15 | 300

4 16 | 1024

5 17 | 306

6 109 | 3379


























[5] Claim for all n 9 | 72n+1+ 22n+1






[7] Claim for all n 13 | 72n+1+ 62n+1










True for n=k rArr 3| _____________






[11] Claim for all n 7 | 8n + 14n + 6



To prove __________________

True for n=k rArr ______________________


rArr 7 | ________________________________


[12] Claim for all n 6 | 7n + 12n + 5


71+12times1 + 5 = 24 and 6|24





2

f(n) = (2n + 1)7n ndash 1


3 Given that

f(n) = 34n + 24n+2




(Total 8 marks)




Claim

Proof by induction



To prove __________________________________________

True for n=k rArr ______________________________________

+ ___________________________

so


= _______________________________________________________



Claim



To prove __________________________________________

True for n=k rArr ______________________________________

+ _________________________________

so


= _______________________________________________________





n

r nn

rr1 1)1(1

(5)


3112

1

2

n

r

nnnr

(5)

3 Show that

n

r

r ndashr1

2 )1ndash ( = 3

1

(n ndash 2)n(n + 2)(6)


12161

1

2

nnnrn

r


22

1

3 )1(41

nnrn

r

(5)



So 2 | 6 3| 6 but 4 | 6 and 5 | 6



31 | 3379 + 31 in other words 31 | 3410





1 11 | 110 rArr 11 | 143 = 110 + 3 times 11

2 7 | 60 + 3 rArr 7 | (60 + 3) times 10 = 630

3 7 | 60 + 3 rArr 7 | 60 + 3times10 = 90


1 13 | 91

2 14 | 210

3 15 | 300

4 16 | 1024

5 17 | 306

6 109 | 3379


























[5] Claim for all n 9 | 72n+1+ 22n+1






[7] Claim for all n 13 | 72n+1+ 62n+1










True for n=k rArr 3| _____________






[11] Claim for all n 7 | 8n + 14n + 6



To prove __________________

True for n=k rArr ______________________


rArr 7 | ________________________________


[12] Claim for all n 6 | 7n + 12n + 5


71+12times1 + 5 = 24 and 6|24





2

f(n) = (2n + 1)7n ndash 1


3 Given that

f(n) = 34n + 24n+2




(Total 8 marks)




Claim



To prove __________________________________________

True for n=k rArr ______________________________________

+ _________________________________

so


= _______________________________________________________





n

r nn

rr1 1)1(1

(5)


3112

1

2

n

r

nnnr

(5)

3 Show that

n

r

r ndashr1

2 )1ndash ( = 3

1

(n ndash 2)n(n + 2)(6)


12161

1

2

nnnrn

r


22

1

3 )1(41

nnrn

r

(5)



So 2 | 6 3| 6 but 4 | 6 and 5 | 6



31 | 3379 + 31 in other words 31 | 3410





1 11 | 110 rArr 11 | 143 = 110 + 3 times 11

2 7 | 60 + 3 rArr 7 | (60 + 3) times 10 = 630

3 7 | 60 + 3 rArr 7 | 60 + 3times10 = 90


1 13 | 91

2 14 | 210

3 15 | 300

4 16 | 1024

5 17 | 306

6 109 | 3379


























[5] Claim for all n 9 | 72n+1+ 22n+1






[7] Claim for all n 13 | 72n+1+ 62n+1










True for n=k rArr 3| _____________






[11] Claim for all n 7 | 8n + 14n + 6



To prove __________________

True for n=k rArr ______________________


rArr 7 | ________________________________


[12] Claim for all n 6 | 7n + 12n + 5


71+12times1 + 5 = 24 and 6|24





2

f(n) = (2n + 1)7n ndash 1


3 Given that

f(n) = 34n + 24n+2




(Total 8 marks)






So 2 | 6 3| 6 but 4 | 6 and 5 | 6



31 | 3379 + 31 in other words 31 | 3410





1 11 | 110 rArr 11 | 143 = 110 + 3 times 11

2 7 | 60 + 3 rArr 7 | (60 + 3) times 10 = 630

3 7 | 60 + 3 rArr 7 | 60 + 3times10 = 90


1 13 | 91

2 14 | 210

3 15 | 300

4 16 | 1024

5 17 | 306

6 109 | 3379


























[5] Claim for all n 9 | 72n+1+ 22n+1






[7] Claim for all n 13 | 72n+1+ 62n+1










True for n=k rArr 3| _____________






[11] Claim for all n 7 | 8n + 14n + 6



To prove __________________

True for n=k rArr ______________________


rArr 7 | ________________________________


[12] Claim for all n 6 | 7n + 12n + 5


71+12times1 + 5 = 24 and 6|24





2

f(n) = (2n + 1)7n ndash 1


3 Given that

f(n) = 34n + 24n+2




(Total 8 marks)





























[5] Claim for all n 9 | 72n+1+ 22n+1






[7] Claim for all n 13 | 72n+1+ 62n+1










True for n=k rArr 3| _____________






[11] Claim for all n 7 | 8n + 14n + 6



To prove __________________

True for n=k rArr ______________________


rArr 7 | ________________________________


[12] Claim for all n 6 | 7n + 12n + 5


71+12times1 + 5 = 24 and 6|24





2

f(n) = (2n + 1)7n ndash 1


3 Given that

f(n) = 34n + 24n+2




(Total 8 marks)




[7] Claim for all n 13 | 72n+1+ 62n+1










True for n=k rArr 3| _____________






[11] Claim for all n 7 | 8n + 14n + 6



To prove __________________

True for n=k rArr ______________________


rArr 7 | ________________________________


[12] Claim for all n 6 | 7n + 12n + 5


71+12times1 + 5 = 24 and 6|24





2

f(n) = (2n + 1)7n ndash 1


3 Given that

f(n) = 34n + 24n+2




(Total 8 marks)








True for n=k rArr 3| _____________






[11] Claim for all n 7 | 8n + 14n + 6



To prove __________________

True for n=k rArr ______________________


rArr 7 | ________________________________


[12] Claim for all n 6 | 7n + 12n + 5


71+12times1 + 5 = 24 and 6|24





2

f(n) = (2n + 1)7n ndash 1


3 Given that

f(n) = 34n + 24n+2




(Total 8 marks)








[11] Claim for all n 7 | 8n + 14n + 6



To prove __________________

True for n=k rArr ______________________


rArr 7 | ________________________________


[12] Claim for all n 6 | 7n + 12n + 5


71+12times1 + 5 = 24 and 6|24





2

f(n) = (2n + 1)7n ndash 1


3 Given that

f(n) = 34n + 24n+2




(Total 8 marks)




[12] Claim for all n 6 | 7n + 12n + 5


71+12times1 + 5 = 24 and 6|24





2

f(n) = (2n + 1)7n ndash 1


3 Given that

f(n) = 34n + 24n+2




(Total 8 marks)







2

f(n) = (2n + 1)7n ndash 1


3 Given that

f(n) = 34n + 24n+2




(Total 8 marks)




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