(1+mx^2)^n = 1+14x^2+21m^2x^4?
2010-10-30 11:05 pm
Given(1+mx^2)^n = 1+14x^2+21m^2x^4+terms involving higher powers of x, find the values of m and n if n is a positive integer.
回答 (1)
2010-10-30 11:47 pm
✔ 最佳答案
(1+mx^2)^n= 1 + nmx^2 + n(n-1)/2 * (mx^2)^2 + ...
= 1 + nmx^2 + n(n-1)/2 * m^2 x^4 + ...
Comparing coefficients,
n(n-1)/2 = 21
n^2 - n - 42 = 0
(n-7)(n+6) = 0
n = 7 or n = -6 (rejected)
and nm = 14
7m = 14
m = 2
參考: Knowledge is power.
收錄日期: 2021-04-25 17:03:19
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20101030000051KK00785