Question

等比数列(n×2*n),q=2,求前n项和

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