Question

裂项相消求详细过程

Answer 1

详见下图：

Answer 2

let
S= 1.3^1+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) -(3^1+3^2+...+3^n)
=n.3^(n+1) -(3/2)(3^n-1)
S=(1/2)n.3^(n+1) -(3/4)(3^n-1)
an
=(4n+1).3^(n+1)
=12(n.3^n) + 3^(n+1)
Tn
=a1+a2+...+an
=12S + (9/2)(3^n -1)
=12[(1/2)n.3^(n+1) -(3/4)(3^n-1)] + (9/2)(3^n -1)
=6n.3^(n+1) -9(3^n-1)+ (9/2)(3^n -1)
=6n.3^(n+1) -(9/2)(3^n-1)
=9/2 + (18n - 9/2).3^n