Binaomial Expansion Nov 27 09
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Transcript of Binaomial Expansion Nov 27 09
The Binomial Theorem, Fibonacci, and Vitruvian Dan
Vitruvian Dan by flickr user nogoodreason
Multiply each of the following
(x + y)2 =
(x + y)3 =
(x + y)4 =
(x + y)5 =
(x + y)0 =
(x + y)1 =
11 1
1 2 11 3 3 1
1 4 6 4 1
What's the pattern?
1. The sum of the exponents is always equal to the power of the binomial.
2. The exponent of the first term begins with the same value as the power of the binomial and decreases by one in each succesive term.
3. The exponent of the second term appears in the second term of the expansion and increases by one until it matches the power of the binomial.
4. The number of terms is one more than the power of the binomial.
5. The coefficients are the combinations of the power number beginning with C(n, 0) and ending at C(n, n) and are symmetrical.
(x + y)n = nC0 xny0 + nC1 xn1y1 + nC2 xn2y2 + . . . . . + nCn x0yn
Evaluate each term ...
(x + y)7 =
(2x y)4 =
Any individual term, let's say the ith term, in a binomial expansion can be represented like this:
ti nC(i 1)an (i 1)b(i 1)=
Find the 4th term in the expansion of
(2 + x)7
x2 2x( )
8Find the 5th term
Determine the indicated term in each expansion.
the 8th term in the expansion of
x 2( )10
Find the term that contains x7 in the expansion of
x3 1x( )9
x3 1x( )
10
Find the term that contains x18 in the expansion of
Try some more
Exercise 34, questions 1 8
Attachments
test by flickr user foreversouls
Vitruvian Dan by flickr user nogoodreason