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SEQUENCE AND SERIES
WORKSHEET SOLUTIONS
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PROBLEM 1
• The sum of 27 consecutive integers is
2646. What is the largest of those
integers?
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PROBLEM 1
• The sum of 27 consecutive integers is
2646. What is the largest of those
integers?• In any arithmetic sequence, the average
is the median.
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PROBLEM 1
• The sum of 27 consecutive integers is
2646. What is the largest of those
integers?• In any arithmetic sequence, the average
is the median.
• Median = 2646 / 27 = 98.
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PROBLEM 1
• The sum of 27 consecutive integers is
2646. What is the largest of those
integers?• In any arithmetic sequence, the average
is the median.
• Median = 2646 / 27 = 98.
• There are 13 numbers above the median.
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PROBLEM 1
• The sum of 27 consecutive integers is
2646. What is the largest of those
integers?• In any arithmetic sequence, the average
is the median.
• Median = 2646 / 27 = 98.
• There are 13 numbers above the median.
• Answer. 98 + 13 = 111
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PROBLEM 2
• If a, b, c, d are positive real numbers such that a,
b, c, d form an increasing arithmetic sequence
and a, b, d form a geometric sequence, find a/d.
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PROBLEM 2
• If a, b, c, d are positive real numbers such that a,
b, c, d form an increasing arithmetic sequence
and a, b, d form a geometric sequence, find a/d.
• We have b = a + Δ, c = a + 2 Δ, and d = a + 3 Δ,
where Δ is a positive real number.
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PROBLEM 2
• If a, b, c, d are positive real numbers such that a,
b, c, d form an increasing arithmetic sequence
and a, b, d form a geometric sequence, find a/d.
• We have b = a + Δ, c = a + 2 Δ, and d = a + 3 Δ,
where Δ is a positive real number.
• Also, b2 = ad yields (a + Δ)2 = a(a + 3 Δ)
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PROBLEM 2
• If a, b, c, d are positive real numbers such that a,
b, c, d form an increasing arithmetic sequence
and a, b, d form a geometric sequence, find a/d.
• We have b = a + Δ, c = a + 2 Δ, and d = a + 3 Δ,
where Δ is a positive real number.
• Also, b2 = ad yields (a + Δ)2 = a(a + 3 Δ)
• Δ2 = a Δ
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PROBLEM 2
• If a, b, c, d are positive real numbers such that a,b, c, d form an increasing arithmetic sequenceand a, b, d form a geometric sequence, find a/d.
• We have b = a + Δ, c = a + 2 Δ, and d = a + 3 Δ,where Δ is a positive real number.
• Also, b2 = ad yields (a + Δ)2 = a(a + 3 Δ)
• Δ2 = a Δ• Δ = a, so the sequence is a, 2a, 3a, 4a, ….
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PROBLEM 2
• If a, b, c, d are positive real numbers such that a,
b, c, d form an increasing arithmetic sequence
and a, b, d form a geometric sequence, find a/d.
• We have b = a + Δ, c = a + 2 Δ, and d = a + 3 Δ,
where Δ is a positive real number.
• Also, b2 = ad yields (a + Δ)2 = a(a + 3 Δ)
• Δ2 = a Δ
• Δ = a, so the sequence is a, 2a, 3a, 4a, ….
• Answer: a/d = 1/4.
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PROBLEM 3
• The sum of 18 consecutive positive
integers is a perfect square. Find the
smallest possible value of this sum.
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PROBLEM 3
• The sum of 18 consecutive positive
integers is a perfect square. Find the
smallest possible value of this sum.• Let the first of the 18 integers be f.
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PROBLEM 3
• The sum of 18 consecutive positive
integers is a perfect square. Find the
smallest possible value of this sum.• Let the first of the 18 integers be f.
• Sum = 18(f+f+17)/ 2 = 9(2f+17)
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PROBLEM 3
• The sum of 18 consecutive positive
integers is a perfect square. Find the
smallest possible value of this sum.• Let the first of the 18 integers be f.
• Sum = 18(f+f+17)/ 2 = 9(2f+17)
• 2f + 17 must be a perfect square.
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PROBLEM 3
• The sum of 18 consecutive positive
integers is a perfect square. Find the
smallest possible value of this sum.• Let the first of the 18 integers be f.
• Sum = 18(f+f+17)/ 2 = 9(2f+17)
• 2f + 17 must be a perfect square.
• Smallest such f is 4.
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PROBLEM 3
• The sum of 18 consecutive positive
integers is a perfect square. Find the
smallest possible value of this sum.• Let the first of the 18 integers be f.
• Sum = 18(f+f+17)/ 2 = 9(2f+17)
• 2f + 17 must be a perfect square.
• Smallest such f is 4.
• Answer: Sum is 9(25) = 225.
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PROBLEM 4
• For all positive integers n less than 2002, let – an = 11, if n is divisible by 13 and 14;
– an = 13, if n is divisible by 14 and 11;
– an = 14, if n is divisible by 11 and 13;
– an = 0, otherwise.• Calculate ∑
=
2001
1n
na
∑=
2001
1n
na
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PROBLEM 4
• For all positive integers n less than 2002, let – an = 11, if n is divisible by 13 and 14;
– an = 13, if n is divisible by 14 and 11;
– an = 14, if n is divisible by 11 and 13;
– an = 0, otherwise.• Calculate
• Notice 2002 = 11(13)(14)
∑=
2001
1n
na
∑=
2001
1n
na
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PROBLEM 4
• For all positive integers n less than 2002, let – an = 11, if n is divisible by 13 and 14;
– an = 13, if n is divisible by 14 and 11;
– an = 14, if n is divisible by 11 and 13;
– an = 0, otherwise.• Calculate
• Notice 2002 = 11(13)(14)
• n < 2002 so
∑=
2001
1n
na
∑=
2001
1n
na
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PROBLEM 4
• For all positive integers n less than 2002, let – an = 11, if n is divisible by 13 and 14;
– an = 13, if n is divisible by 14 and 11;
– an = 14, if n is divisible by 11 and 13;
– an = 0, otherwise.• Calculate
• Notice 2002 = 11(13)(14)
• n < 2002 so• There are ten 11s, twelve 13s, and thirteen 14s.
∑=
2001
1n
na
∑=
2001
1n
na
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PROBLEM 4
• For all positive integers n less than 2002, let – an = 11, if n is divisible by 13 and 14;
– an = 13, if n is divisible by 14 and 11;
– an = 14, if n is divisible by 11 and 13;
– an = 0, otherwise.• Calculate
• Notice 2002 = 11(13)(14)
• n < 2002 so• There are ten 11s, twelve 13s, and thirteen 14s.
• The sum is 10(11) + 12(13) + 13(14)
• Answer 448.
∑=
2001
1n
na
∑=
2001
1n
na
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PROBLEM 5
• Let {ak} be a sequence of integers suchthat a1 = 1 and am+n = am +an +mn,for allpositive integers m and n. Find a12.
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PROBLEM 5
• Let {ak} be a sequence of integers suchthat a1 = 1 and am+n = am +an +mn,for all
positive integers m and n. Find a12.
• Consider the case when n=1.
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PROBLEM 5
• Let {ak} be a sequence of integers suchthat a1 = 1 and am+n = am +an +mn,for all
positive integers m and n. Find a12.
• Consider the case when n=1.
• am+1 = am + 1 + (m)(1) = am + (m+1)
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PROBLEM 5
• Let {ak} be a sequence of integers suchthat a1 = 1 and am+n = am +an +mn,for all
positive integers m and n. Find a12.
• Consider the case when n=1.
• am+1 = am + 1 + (m)(1) = am + (m+1)
• am = 1 + 2 + … + m
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PROBLEM 5
• Let {ak} be a sequence of integers suchthat a1 = 1 and am+n = am +an +mn,for all
positive integers m and n. Find a12.
• Consider the case when n=1.
• am+1 = am + 1 + (m)(1) = am + (m+1)
• am = 1 + 2 + … + m
• am = m(m+1)/2
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PROBLEM 5
• Let {ak} be a sequence of integers suchthat a1 = 1 and am+n = am +an +mn,for all
positive integers m and n. Find a12.
• Consider the case when n=1.
• am+1 = am + 1 + (m)(1) = am + (m+1)
• am = 1 + 2 + … + m
• am = m(m+1)/2
• a12 = 12(13)/2 = 78
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PROBLEM 6
• Let a1, a2, . . . be a sequence for which a1 = 2,a2 = 3, and a
n= a
n-1/a
n-2for each positive integer
n > 2. What is a2006?
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PROBLEM 6
• Let a1, a2, . . . be a sequence for which a1 = 2,a2 = 3, and a
n= a
n-1/a
n-2for each positive integer
n > 2. What is a2006?
• We cannot expect to find all terms from 1 to2006 in a reasonable length of time.
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PROBLEM 6
• Let a1, a2, . . . be a sequence for which a1 = 2,a2 = 3, and a
n= a
n-1/a
n-2for each positive integer
n > 2. What is a2006?
• We cannot expect to find all terms from 1 to2006 in a reasonable length of time.
• We must look for a pattern.
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PROBLEM 6
• Let a1, a2, . . . be a sequence for which a1 = 2,a2 = 3, and a
n= a
n-1/a
n-2for each positive integer
n > 2. What is a2006?
• We cannot expect to find all terms from 1 to2006 in a reasonable length of time.
• We must look for a pattern.
• a3= 3/2, a4= 1/2, a5 = 1/3, a6 = 2/3, a7 = 2, a8 = 3.
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PROBLEM 6
• Let a1, a2, . . . be a sequence for which a1 = 2,a2 = 3, and a
n= a
n-1/a
n-2for each positive integer
n > 2. What is a2006?
• We cannot expect to find all terms from 1 to2006 in a reasonable length of time.
• We must look for a pattern.
• a3= 3/2, a4= 1/2, a5 = 1/3, a6 = 2/3, a7 = 2, a8 = 3.• The sequence is periodic with a period of 6,
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PROBLEM 6
• Let a1, a2, . . . be a sequence for which a1 = 2,a2 = 3, and a
n= a
n-1/a
n-2for each positive integer
n > 2. What is a2006?
• We cannot expect to find all terms from 1 to2006 in a reasonable length of time.
• We must look for a pattern.
• a3= 3/2, a4= 1/2, a5 = 1/3, a6 = 2/3, a7 = 2, a8 = 3.
• The sequence is periodic with a period of 6,
• so a2006
= a2
= 3.
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PROBLEM 7
• Consider the sequence of numbers: 4, 7, 1, 8, 9, 7, 6, . . ..For n > 2, the nth term of the sequence is the units digit of
the sum of the two previous terms. Let Sn denote the
sum of the first n terms of this sequence. What is the
smallest value of n for which Sn > 10, 000?
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PROBLEM 7
• Consider the sequence of numbers: 4, 7, 1, 8, 9, 7, 6, . . ..For n > 2, the nth term of the sequence is the units digit of
the sum of the two previous terms. Let Sn denote the
sum of the first n terms of this sequence. What is the
smallest value of n for which Sn > 10, 000?
• Writing out more terms of the sequence yields
4, 7, 1, 8, 9, 7, 6, 3, 9, 2, 1, 3, 4, 7, 1 . . . .
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PROBLEM 7
• Consider the sequence of numbers: 4, 7, 1, 8, 9, 7, 6, . . ..For n > 2, the nth term of the sequence is the units digit of
the sum of the two previous terms. Let Sn denote the
sum of the first n terms of this sequence. What is the
smallest value of n for which Sn > 10, 000?
• Writing out more terms of the sequence yields
4, 7, 1, 8, 9, 7, 6, 3, 9, 2, 1, 3, 4, 7, 1 . . . .
• The sequence repeats itself, starting with the 13th term.
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PROBLEM 7
• Consider the sequence of numbers: 4, 7, 1, 8, 9, 7, 6, . . ..For n > 2, the nth term of the sequence is the units digit of
the sum of the two previous terms. Let Sn denote the
sum of the first n terms of this sequence. What is the
smallest value of n for which Sn > 10, 000?
• Writing out more terms of the sequence yields
4, 7, 1, 8, 9, 7, 6, 3, 9, 2, 1, 3, 4, 7, 1 . . . .
• The sequence repeats itself, starting with the 13th term.• Because S12 = 60, S12k = 60k for all positive integers k.
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PROBLEM 7
• Consider the sequence of numbers: 4, 7, 1, 8, 9, 7, 6, . . ..For n > 2, the nth term of the sequence is the units digit of
the sum of the two previous terms. Let Sn denote the
sum of the first n terms of this sequence. What is the
smallest value of n for which Sn > 10, 000?
• Writing out more terms of the sequence yields
4, 7, 1, 8, 9, 7, 6, 3, 9, 2, 1, 3, 4, 7, 1 . . . .
• The sequence repeats itself, starting with the 13th term.• Because S12 = 60, S12k = 60k for all positive integers k.
• The largest k for which S12k ≤ 10, 000 is
• k = [10, 000/60] = 166, and S12(166) = 60(166) = 9960.
PROBLEM
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PROBLEM 7
• Writing out more terms of the sequence yields4, 7, 1, 8, 9, 7, 6, 3, 9, 2, 1, 3, 4, 7, 1 . . . .
• The sequence repeats itself, starting with the 13th term.
• Because S12 = 60, S12k = 60k for all positive integers k.• The largest k for which S12k ≤ 10, 000 is
• k = [10, 000/60] = 166, and S12(166) = 60(166) = 9960.
PROBLEM 7
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PROBLEM 7
• Writing out more terms of the sequence yields4, 7, 1, 8, 9, 7, 6, 3, 9, 2, 1, 3, 4, 7, 1 . . . .
• The sequence repeats itself, starting with the 13th term.
• Because S12 = 60, S12k = 60k for all positive integers k.• The largest k for which S12k ≤ 10, 000 is
• k = [10, 000/60] = 166, and S12(166) = 60(166) = 9960.
• To have Sn > 10, 000, add enough additional terms for their sum to exceed 40.
PROBLEM 7
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PROBLEM 7
• Writing out more terms of the sequence yields4, 7, 1, 8, 9, 7, 6, 3, 9, 2, 1, 3, 4, 7, 1 . . . .
• The sequence repeats itself, starting with the 13th term.
• Because S12 = 60, S12k = 60k for all positive integers k.• The largest k for which S12k ≤ 10, 000 is
• k = [10, 000/60] = 166, and S12(166) = 60(166) = 9960.
• To have Sn > 10, 000, add enough additional terms for their sum to exceed 40.
• This can be done by adding the next 7 terms of the
sequence, since their sum is 42.
PROBLEM 7
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PROBLEM 7
• Writing out more terms of the sequence yields4, 7, 1, 8, 9, 7, 6, 3, 9, 2, 1, 3, 4, 7, 1 . . . .
• The sequence repeats itself, starting with the 13th term.
• Because S12 = 60, S12k = 60k for all positive integers k.• The largest k for which S12k ≤ 10, 000 is
• k = [10, 000/60] = 166, and S12(166) = 60(166) = 9960.
• To have Sn > 10, 000, add enough additional terms for their sum to exceed 40.
• This can be done by adding the next 7 terms of the
sequence, since their sum is 42.
• Thus, the smallest value of n is 12(166) + 7 = 1999.
PROBLEM 8
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PROBLEM 8
• Suppose that {an} is an arithmetic sequence with• a1 + a2 +・・・ + a100 = 100 and
• a101 + a102 +・・・ + a200 = 200.
• What is the value of a2−
a1?
PROBLEM 8
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PROBLEM 8
• Suppose that {an} is an arithmetic sequence with• a1 + a2 +・・・ + a100 = 100 and
• a101 + a102 +・・・ + a200 = 200.
• What is the value of a2−
a1?• Let Δ = a2 − a1
PROBLEM 8
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PROBLEM 8
• Suppose that {an} is an arithmetic sequence with• a1 + a2 +・・・ + a100 = 100 and
• a101 + a102 +・・・ + a200 = 200.
• What is the value of a2−
a1?• Let Δ = a2 − a1
• a101 − a1 = 100 Δ = a100+ n − an for all n
PROBLEM 8
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PROBLEM 8
• Suppose that {an} is an arithmetic sequence with• a1 + a2 +・・・ + a100 = 100 and
• a101 + a102 +・・・ + a200 = 200.
• What is the value of a2−
a1?• Let Δ = a2 − a1
• a101 − a1 = 100 Δ = a100+ n − an for all n
• (a101 + a102 +・・・ + a200) = 200
• – (a1 + a2 +・・・ + a100) = 100
PROBLEM 8
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• Suppose that {an} is an arithmetic sequence with• a1 + a2 +・・・ + a100 = 100 and
• a101 + a102 +・・・ + a200 = 200.
• What is the value of a2−
a1?• Let Δ = a2 − a1
• a101 − a1 = 100 Δ = a100+ n − an for all n
• (a101 + a102 +・・・ + a200) = 200
• – (a1 + a2 +・・・ + a100) = 100
• 100(100 Δ) = 100
PROBLEM 8
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• Suppose that {an} is an arithmetic sequence with• a1 + a2 +・・・ + a100 = 100 and
• a101 + a102 +・・・ + a200 = 200.
• What is the value of a2−
a1?• Let Δ = a2 − a1
• a101 − a1 = 100 Δ = a100+ n − an for all n
• (a101 + a102 +・・・ + a200) = 200
• – (a1 + a2 +・・・ + a100) = 100
• 100(100 Δ) = 100
• Δ = 1/100 = a2 − a1
PROBLEM 9
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• Given a finite sequence S = (a1, a2, . . ., an) of nreal numbers, let A(S) be the sequence
( (a1 + a2)/2, (a2 + a3)/2, . . ., (an-1 + an)/2) of n-1
real numbers. Define A1
(S
) = A
(S
) and, for eachinteger m, 1 < m < n, define Am(S) = A( Am-1(S)).
Suppose x > 0, and let S = (1, x, x2, . . ., x100). If
A100(S) = 1/250, then what is x ?
PROBLEM 9
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• Given a finite sequence S = (a1, a2, . . ., an) of nreal numbers, let A(S) be the sequence
( (a1 + a2)/2, (a2 + a3)/2, . . ., (an-1 + an)/2) of n-1
real numbers. Define A1
(S
) = A
(S
) and, for eachinteger m, 1 < m < n, define Am(S) = A( Am-1(S)).
Suppose x > 0, and let S = (1, x, x2, . . ., x100). If
A100(S) = 1/250, then what is x ?
• It is clear that the denominator of each term of
Am
(S
) = 2m
.
PROBLEM 9
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• Given a finite sequence S = (a1, a2, . . ., an) of nreal numbers, let A(S) be the sequence
( (a1 + a2)/2, (a2 + a3)/2, . . ., (an-1 + an)/2) of n-1
real numbers. Define A1
(S
) = A
(S
) and, for eachinteger m, 1 < m < n, define Am(S) = A( Am-1(S)).
Suppose x > 0, and let S = (1, x, x2, . . ., x100). If
A100(S) = 1/250, then what is x ?
• It is clear that the denominator of each term of
Am
(S
) = 2m
.• Let’s investigate the numerators of each term.
PROBLEM 9
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• S is 1, x, x2, x3, x4, …
PROBLEM 9
7/28/2019 Sequences and Series Solutions
http://slidepdf.com/reader/full/sequences-and-series-solutions 55/66
• S is 1, x, x2, x3, x4, …• The numerators of A1(S) = 1+x, x+x2, x2+x3, …
PROBLEM 9
7/28/2019 Sequences and Series Solutions
http://slidepdf.com/reader/full/sequences-and-series-solutions 56/66
• S is 1, x, x2, x3, x4, …• The numerators of A1(S) = 1+x, x+x2, x2+x3, …
• The numerators of A2(S) = 1+2x+x2, x+2x2+x3,…
PROBLEM 9
7/28/2019 Sequences and Series Solutions
http://slidepdf.com/reader/full/sequences-and-series-solutions 57/66
• S is 1, x, x2, x3, x4, …• The numerators of A1(S) = 1+x, x+x2, x2+x3, …
• The numerators of A2(S) = 1+2x+x2, x+2x2+x3,…
• The numerators of A3(S) = 1+3x+3x2+x3, …
PROBLEM 9
7/28/2019 Sequences and Series Solutions
http://slidepdf.com/reader/full/sequences-and-series-solutions 58/66
• S is 1, x, x2, x3, x4, …• The numerators of A1(S) = 1+x, x+x2, x2+x3, …
• The numerators of A2(S) = 1+2x+x2, x+2x2+x3,…
• The numerators of A3(S) = 1+3x+3x2+x3, …
• The numerator of the 1st term of Am(S) = (1+x)m
PROBLEM 9
7/28/2019 Sequences and Series Solutions
http://slidepdf.com/reader/full/sequences-and-series-solutions 59/66
• S is 1, x, x2
, x3
, x4
, …• The numerators of A1(S) = 1+x, x+x2, x2+x3, …
• The numerators of A2(S) = 1+2x+x2, x+2x2+x3,…
• The numerators of A3(S) = 1+3x+3x2+x3, …
• The numerator of the 1st term of Am(S) = (1+x)m
• Am(S) has one fewer term than Am-1(S) and
A1(S) has 100 terms
PROBLEM 9
7/28/2019 Sequences and Series Solutions
http://slidepdf.com/reader/full/sequences-and-series-solutions 60/66
• S is 1, x, x2
, x3
, x4
, …• The numerators of A1(S) = 1+x, x+x2, x2+x3, …
• The numerators of A2(S) = 1+2x+x2, x+2x2+x3,…
• The numerators of A3(S) = 1+3x+3x2+x3, …
• The numerator of the 1st term of Am(S) = (1+x)m
• Am(S) has one fewer term than Am-1(S) and
A1(S) has 100 terms
• A100(S) has exactly one term.
PROBLEM 9
7/28/2019 Sequences and Series Solutions
http://slidepdf.com/reader/full/sequences-and-series-solutions 61/66
• S is 1, x, x2
, x3
, x4
, …• The numerators of A1(S) = 1+x, x+x2, x2+x3, …
• The numerators of A2(S) = 1+2x+x2, x+2x2+x3,…
• The numerators of A3(S) = 1+3x+3x2+x3, …
• The numerator of the 1st term of Am(S) = (1+x)m
• Am(S) has one fewer term than Am-1(S) and
A1(S) has 100 terms
• A100(S) has exactly one term.
• A100
(S
) = (1 + x)100
/ 2100
= 1 / 250
PROBLEM 9
7/28/2019 Sequences and Series Solutions
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• (1 + x)100
/ 2100
= 1 / 250
PROBLEM 9
7/28/2019 Sequences and Series Solutions
http://slidepdf.com/reader/full/sequences-and-series-solutions 63/66
• (1 + x)100
/ 2100
= 1 / 250
• (1 + x)2 / 22 = 1 / 2
PROBLEM 9
7/28/2019 Sequences and Series Solutions
http://slidepdf.com/reader/full/sequences-and-series-solutions 64/66
• (1 + x)100
/ 2100
= 1 / 250
• (1 + x)2 / 22 = 1 / 2
• (1 + x)2 = 2
PROBLEM 9
7/28/2019 Sequences and Series Solutions
http://slidepdf.com/reader/full/sequences-and-series-solutions 65/66
• (1 + x)100
/ 2100
= 1 / 250
• (1 + x)2 / 22 = 1 / 2
• (1 + x)2 = 2
• 1 + x = √2
PROBLEM 9