Question

已知数列{Cn),Cn=n*2^n求数列{Cn)的前n项和Sn

Answer 1

let

S = 1.2^1 +2.2^2+.+n.2^n (1)

2S = 1.2^2 +2.2^3+.+n.2^(n+1) (2)

(2)-(1)

S= n.2^(n+1) - ( 2+2^2+...+2^n)

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

= 2+ (2n-2).2^n

Sn = S = 2+ (2n-2).2^n