Question

幂级数求和

Answer 1

先看第2个吧
∑都为0到无穷
∑(n-1)^2x^n/(n+1)
=∑[(n+1)^2-4n]x^n/(n+1)
=∑(n+1)x^n-∑4(n+1-1)x^n/(n+1)
=∑(n+1)x^n-4∑x^n+4∑x^n/(n+1)
=[∑x^(n+1)]'-4∑x^n+4[∑x^(n+1)/(n+1)]/x
需要知道的展开式有：1/1-x=∑x^n和ln(1-x)=-∑x^(n+1)/(n+1)
所以[∑x^(n+1)]'-4∑x^n+4[∑x^(n+1)/(n+1)]/x
=[x/1-x]'-4x-4ln(1-x)/x
=[(4x-3)/(x-1)^2]-4ln(1-x)/x
第一个太简单了直接分开就是了
=2∑nx^n+∑x^n
=2x[∑x^n]'+∑x^n
=2x/(x-1)^2+1/1-x
=(x+1)/(x-1)^2

Answer 2

1.(x+1)/(x-1)^2
2.[(4x-3)/(x-1)^2]-4ln(1-x)/x