cn=(2n+1)8∧(n-1)用错位相减法求和

let
S =1.8^0+2.8^1+.+n.8^(n-1) (1)
8S = 1.8^1+2.8^2+.+n.8^n (2)
(2)-(1)
7S = n.8^n -[1+8+...+8^(n-1) ]
=n.8^n -(1/7)(8^n -1)
S = (1/7)[ n.8^n -(1/7)(8^n -1) ]
cn= (2n+1).8^(n-1)
= 2[n.8^(n-1)] + 8^(n-1)
Sn = c1+c2+...+cn
=2S + (1/7)(8^n -1)
=(2/7)[n.8^n -(1/7)(8^n -1)] +(1/7)(8^n -1)
=(2/7)n.8^n +(5/49)(8^n -1)
= (1/49)[ -5 + (14n+5).8^n ]