Question

简单数列求和

An=(2n-1)4^n-1求这个数列的前N项和，大家帮帮忙，谢啦

Answer 1

a(n)=(2n-1)4^(n-1)
s(n)=(2*1-1)+(2*2-1)4+(2*3-1)4^2+...+[2(n-1)-1]4^(n-2)+(2n-1)4^(n-1)
4s(n)=(2*1-1)4 + (2*2-1)4^2 + (2*3-1)4^3+...+[2(n-1)-1]4^(n-1) + (2n-1)4^n
-3s(n)=s(n)-4s(n)=(2*1-1)+(2*1)4 +(2*1)4^2+...+(2*1)4^(n-1)-(2n-1)4^n
=2[1+4+4^2+...+4^(n-1)] - 1 - (2n-1)4^n
=2[4^n-1]/3 - 1 - (2n-1)4^n
=[5/3-2n]4^n - 5/3
s(n)=[6n-5]4^n/9 + 5/9