Philip P. answered • 04/01/14

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You already asked this question and the answer was provided:

http://www.wyzant.com/resources/answers/30927/which_term_of_the_expansion_contains_x_8

Is there something additional you need?

Philip P.

*In your case, (x + 5)*

^{10}, a=x, b=5, n=10. What do you have to set k to to get x^{n-k}= x8? The term number, then is r = k+1.OK, to make x

^{10-k}= x^{8}, k must = 2. The term number is r = k +1 = 2 + 1 = 3. Is that what you got?
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04/01/14

Philip P.

The actual term itself (vice just the number of the term) can be determined from the following formula:

(n!/(n-k)!k!)*a

^{n-k}bk = 10!/(10-2)!(2!) * x

^{8}5

^{2 }= 1125x

^{8}

Where n = 10, k = 2, a = x, b = 5.

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04/01/14

Dalia S.

04/01/14