Question

an=(3n-2)(1/2)^(n-1),求{an}前n项和

Answer 1

Sn=a1+a2+a3+...+an
=(3-2)(1/2)^0+(6-2)(1/2)^1+(9-3)(1/2)^2+...+(3n-5)(1/2)^(n-2)+(3n-2)(1/2)^(n-1)

Sn/2
= (3-2)(1/2)^1+ (6-2)(1/2)^2+...+(3n-8)(1/2)^(n-2)+(3n-5)(1/2)^(n-1)+(3n-2)(1/2)^n
两式相减得:
Sn/2
=1+3*(1/2)^1+3*(1/2)^2+...+3*(1/2)^(n-1)-(3n-2)(1/2)^n
=1+3*(1-(1/2)^(n-1))-(3n-2)(1/2)^n
=1+3-3*(1/2)^(n-1)-3n(1/2)^n+(1/2)^(n-1)
=4-(1/2)^(n-2)-3n(1/2)^n
所以
Sn=8-(1/2)^(n-3)-3n(1/2)^(n-1)