已知数列{an}中,an=(3n-2)•3的n+1次方,求Sn?

已知数列{an}中,an=(3n-2)•3的n+1次方,求Sn?
已知数列{an}中,an=(3n-2)•3的n+1次方,求Sn?

已知数列{an}中,an=(3n-2)•3的n+1次方,求Sn?
an = (3n-2).3^(n+1)
= 9(n.3^n) -2.3^(n+1)
Sn = an+a2+...+an
=9[∑(i:1->n)i.3^i ] - 9(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 =n.3^(n+1) -( 3+3^2+...+3^n)
=n.3^(n+1) - (3/2)(3^n-1)
S = (1/2)[n.3^(n+1) - (3/2)(3^n-1)]
Sn = an+a2+...+an
=9[∑(i:1->n)i.3^i ] - 9(3^n-1)
=9S -9(3^n-1)
= (9/2)[n.3^(n+1) - (3/2)(3^n-1)] -9(3^n-1)
= (9/2)n.3^(n+1) - (7/4) .3^(n+2) + 9

an=（9n_6）乘（9n_6）的n次方，然后用等比数列求和公式