已知数列an的通项公式为an等于(3n十1）乘三的n次方求数列前n项和

Question

如题所述

Answer 1

an = (3n+1).3^n
= 3(n.3^n) + 3^n
Sn =a1+a2+...+an
=3[∑(i:1->n) i.3^i] + (3/2)(3^n-1)
let
S = 1.3 +2.3^2+...+n.3^n (1)
3S = 1.3^2 +2.3^3+...+n.3^(n+1) (2)
(2)-(1)
2S = n3^(n+1) - ( 3+3^2+...+3^n)
=n3^(n+1) - (3/2)(3^n-1)
S = 2n.3^(n+1) - 3(3^n-1)

Sn =a1+a2+...+an
=3[∑(i:1->n) i.3^i] + (3/2)(3^n-1)
=3[2n.3^(n+1) - 3(3^n-1) ] +(3/2)(3^n-1)
= 15/2 + (3/2)(12n-5).3^n