if the coefficent of x^3 is triple that of x in the expansion of (1- x/2) ^ n ,where n is a positive integer ,find the value of n.
how to do?
maths m1
2011-09-24 9:06 pm
回答 (1)
2011-09-24 9:33 pm
✔ 最佳答案
Coefficient of x = n(-1/2)Coefficient of x^3 = n(n - 1)(n - 2)/3! [-1/2]^3 = -n(n - 1)(n - 2)/48
So 3n(-1/2) = -n(n - 1)(n - 2)/48
72 = (n - 1)(n - 2)
n^2 - 3n - 70 = 0
(n + 7)(n - 10) = 0
n = - 7 (rejected since n is a positive integer) or 10
so n = 10.
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