F.4 @ Binomial Theorem
2010-01-06 4:03 am
In the expansion of [3(x^2) – (1/x)]^n, the 5th term in descending powers of x is the constant term, where n is a positive integer. Find the value of n and the 5th term.
回答 (1)
2010-01-06 4:14 am
✔ 最佳答案
(3x^2 - 1/x)^nThe general term, r + 1 th term
= nCr (3x^2)^n-r (-1/x)^r
= nCr (-1)^r(3)^n-r x^(2n - 3r)
The 5th term is a constant term,
2n - 3(4) = 0
n = 6
The 5th term
= 6C4 (-1)^4(3)^6-4
= 135
參考: Physics king
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