It is given that [(4x^2)+3]^n = a+(bx^2)+(cx^4)+(dx^6)+terms involving higher powers of x. If 448b=21d and n is a positive integer, find the value of n.
ans：n=10
don't skip steps!!
binomial theorem
2010-12-23 3:08 am
回答 (1)
2010-12-23 3:19 am
✔ 最佳答案
[3+(4x^2)]^n = a+(bx^2)+(cx^4)+(dx^6)+terms involving higher powers of x.(3^n)+n[3^(n-1)](4x^2)+(nC2)[3^(n-2)](16x^4)+(nC3)[3^(n-3)](64x^6)
=a+(bx^2)+(cx^4)+(dx^6)+terms involving higher powers of x.
Comparing the coefficients
b=4n[3^(n-1)],d=64(nC3)[3^(n-3)]
Since 448b=21d
1792n[3^(n-1)]=1344(nC3)[3^(n-3)]
16128n=1344(nC3)
12n=nC3=n(n-1)(n-2)/6
72=(n-1)(n-2)
n=10
收錄日期: 2021-04-19 23:44:39
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20101222000051KK00855