[2x+1/(2x)]^2n的展开式中x^2的系数224,求(1/x)^2系数是?
问题描述：
[2x+1/(2x)]^2n的展开式中x^2的系数224,求(1/x)^2系数是?
答
(a+b)^3=a^3+3a^2b+3ab^2+b^3
(a+b)^2=a^2+2ab+b^2
原式=(4x^2+2+1/4x^2)^n
若n=2k 则a^2=m(4x^2)^2=16mx^2=224x^2,m=14
2ab=2m(4x^2+2)*(1/4x^2)=2m+m/x^2 系数=14
若n=3k 则3a^2b+3ab^2+b^3=m(51x^2+15/4x^2),m≠k
综上 =14