Question

1.求数列{n*2^n-n}的前n项和

Answer 1

1.设Sn=n2^n-n 由错位相减法得

Sn=1*2^1+2*2^2+3*2^3+...+n*2^n-(1+2+.+n)①

2Sn= 1*2^2+2*2^3+...+(n-1)*2^n+n*2^(n+1)-2(1+2+...+n)②

②-① Sn=n*2^(n+1)-(1+2+...+n)-(2^2+2^3+...+2^n)-2

=(n+1)2^(n+1)-n(n+1)/2+2

2.是（2n-3）/(2^(n-3))么