bn=(2n+1)乘2的(2n+1)次方的前n项和

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) - (2^1+2^2+...+2^n)
=n.2^(n+1) - (2^n -1 )
4S =4n.2^(n+1) - 4(2^n -1 )
bn = (2n+1).2^(n+1)
= 4(n.2^n) + 2^(n+1)
Tn = b1+b2+...+bn
= 4S + 4(2^n-1)
=4n.2^(n+1)
=8n.2^n