首先[]中为等比数列,按等比数列求和公式Sn=[a1*(1-q^n)]/(1-q)可得
1+x+x^2+x^3+…+x^(n-2)]
=[1*(1-x^(n-1))]/(1-x)
=(1-x^(n-1))/(1-x)
继续化简
1+2x[1+x+x^2+x^3+…+x^(n-2)]-(2n-1)*x^n
=1+2x*(1-x^(n-1))/(1-x)-(2n-1)*x^n;
=1+(2x-2x^n)/(1-x)-(2n-1)*x^n
=(1+x-2x^n)/(1-x)-(2n-1)*x^n
通分=[(2n-1)*x^(n+1)-(2n+1)*x^n+(1+x)]/(1-x)
两边再同时除以1-x
则可知
Sn=(2n-1)*x^(n+1)-(2n+1)*x^n+(1+x)/(1-x)^2