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 xn-k = x8? The term number, then is r = k+1.
OK, to make x10-k = x8, 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!)*an-kbk = 10!/(10-2)!(2!) * x852 = 1125x8
Where n = 10, k = 2, a = x, b = 5.
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04/01/14
Dalia S.
04/01/14