Question

数学数列题.在数列an中,a1=1,a(n+1)-(n+1)=2(an-1),bn=an+n(1)求证bn是等比数列(2

Answer 1

(1)

a(n+1) -(n+1) = 2(an -1)

a(n+1) =2an +n-1

a(n+1) + (n+1) = 2(an + n )

{an + n } 是等比数列,q=2

bn = an +n 是等比数列

(2)

an + n = 2^(n-1) .( a1+ 1)

= 2^n

an = -n+2^n

bn = an+n = 2^n

let

S = 1.(1/2)^0 +2.(1/2)^1+.+n.(1/2)^(n-1) (1)

(1/2)S = 1.(1/2)^1 +2.(1/2)^2+.+n.(1/2)^n (2)

(1)-(2)

(1/2)S =[ 1+1/2+.+1/2^(n-1)] - n.(1/2)^n

= 2[1- 1/2^n] - n.(1/2)^n

S = 4[1- 1/2^n] - 2n.(1/2)^n

cn ={(2n-1)/(an+logbn)}

= (2n-1)/2^n

= n(1/2)^(n-1) - 1/2^n

Sn = c1+c2+...+cn

= S - 2(1- 1/2^n)

=2[1- 1/2^n] - 2n.(1/2)^n

= 2 - (2n+2) (1/2)^n