Question

已知数列an的通项an=(2n-1)·3^n-1,求an的前n项和Sn

Answer 1

an=(2n-1)·3^(n-1)

= 2[ n.3^(n-1) ] - 3^(n-1)

Sn = a1+a2+...+an

= 2[∑(i:1->n) i.3^(i-1) ] - (3^n-1)/2

let

S = 1.3^0+ 2.3^1+.+n.3^(n-1) (1)

3S = 1.3^1+ 2.3^2+.+n.3^n (2)

(2)-(1)

2S = n.3^n - [ 3^0+3^1+.+3^(n-1) ]

=n.2^n - (3^n-1)/2

S = 2n.2^n - (3^n-1)

Sn = 2[∑(i:1->n) i.3^(i-1) ] - (3^n-1)/2

= 2S -(3^n-1)/2

= 4n.2^n - (3/2)( 3^n-1)