In the expansion of [x^2 + (k/x) ]^8, where k is not equal to 0, the coeff of x^7 is equal to the coeff of 1/(x^5). Find the values of k.
I want to have the solution.
a.math (binomial theorem)
2006-10-14 9:07 pm
回答 (2)
2006-10-14 9:38 pm
✔ 最佳答案
[x^2+(k/x)]^8= summation of 8Cn (x^2)^(8-n) (k/x)^n where n is integer from 0 to 8
= summation of 8Cn k^n [x^(16-3n)] where n is integer from 0 to 8
When 16-3n = 7
n = 3
When 16-3n = -5
n = 7
Therefore 8C3 k^3 = 8C7 k^7
56 / 8 = k^4
k = 7^0.25
2006-10-14 9:18 pm
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