數學M1 關於(3x^2+1)^n
2010-10-30 9:21 pm
Given(3x^2+1)^n=a+bx^2+cx^4+dx^6+terms involving higher powers of x, find n if d=108b and n is a positive integer.
回答 (1)
2010-10-30 10:09 pm
✔ 最佳答案
(3x^2 + 1)^n= (1 + 3x^2)^n
= 1 + n(3x^2) + n(n-1)/2 * (3x^2)^2 + n(n-1)(n-2)/6 * (3x^2)^3 + ...
= 1 + 3nx^2 + 9n(n-1)/2 * x^4 + 9n(n-1)(n-2)/2 * x^6 + ...
Hence, a = 1
b = 3n
c = 9n(n-1)/2
d = 9n(n-1)(n-2)/2
Since d = 108b
9n(n-1)(n-2)/2 = 108(3n)
9n(n-1)(n-2) = 648n
n^2 - 3n + 2 = 72
n^2 - 3n - 70 = 0
(n-10)(n+7) = 0
n = 10 or n = -7
參考: Knowledge is power.
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