高一数列问题 有加分！！！

则第n项为x^（n-1）+x^n+...+x^（2n-2）
=x^(n-1)[1+x+...+x^(n-1)]
=x^(n-1)*[(1-x^n)/(1-x)]
=1/(1-x)*[x^(n-1)-x^n]
所以通项为：an=1/(1-x)*[x^(n-1)-x^（2n-1）]
则
Sn=∑1/(1-x)*x^(n-1)-∑1/(1-x)*x^（2n-1）
=1/(1-x)*[(1-x^n)/(1-x)]-x/(1-x)*∑（x2）^（n-1）
=(1-x^n)/(1-x)^2-x/(1-x)*[1-（x2）^n]/(1-x^2)
=[(1-x^n)（1+x）-x*（1-（x2）^n）]/[(1-x)^2(1+x）]
=(1-x^n)^2/[(1-x)^2(1+x）]

an=x^(n-1)+x^n+……+x^(2n-2),一共n项
所以an=x^(n-1)*(x^n-1)/(x-1)
=[x^(2n-1)-x^(n-1)]/(x-1)

即求x^(2n-1)的前n项和，再减去x^(n-1)的前n项和，最后除以(x-1)

x^(2n-1)首项x^(2*1-1)=x,公比x^2
所以和=x*(x^2n-1)/(x^2-1)

x^(n-1)首项x^(1-1)=x^0=1,公比x
所以和=1*(x^n-1)/(x-1)

所以Sn=[x*(x^2n-1)/(x^2-1)-(x^n-1)/(x-1)]/(x-1)
=[x*(x^2n-1)/(x^2-1)-(x+1)(x^n-1)/(x^-1)]/(x-1)
=[x^(2n+1)-x-x^(n+1)+x-x^n+1]/(x-1)(x^2-1)
=[x^(2n+1)-x^(n+1)-x^n+1]/(x-1)(x^2-1)
=[x^(n+1)-1](x^n-1)/(x-1)(x^2-1)

=1+x+2(x^2+x^3+....+x^(n-2))+x^n-1+x^n
=1+x+x^n-1+x^n-2+2x^2(1-x^(n-4))/(