幂级数求和

2个回答

  • 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