Question

cn=(2n+1)8∧(n-1)用错位相减法求和

Answer 1

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 ]