已知数列的通项公式是an=3n/2的n-1次方,求数列前n项和Sn

let
S = 1.(1/2)^0+2.(1/2)^1+.+n.(1/2)^(n-1) (1)
(1/2)S = 1.(1/2)^1+2.(1/2)^2+.+n.(1/2)^n (2)
(2)-(1)
(1/2)S = [1+1/2+1/2^2+...+1/2^(n-1) ] -n.(1/2)^n
= ( 1- 1/2^n) -n.(1/2)^n
S = 2[( 1- 1/2^n) -n.(1/2)^n]
an=3n/2^(n-1)
= 3[ n.(1/2)^(n-1) ]
Sn = a1+a2+.+an
= 3S
=6[( 1- 1/2^n) -n.(1/2)^n]
=6 - (6n+6)(1/2)^n