题目

已知数列an满足a1+2a2+...+2^(n-1)*an=n/2,求an通项.若bn=n/an,求bn前n项和Sn

提问时间：2021-01-02

答案

a1+2a2+...+2^(n-1)*an=n/2 (1)
a1+2a2+...+2^(n-2)*a(n-1)=(n-1)/2 (2)
(1)-(2)
2^(n-1)*an=n/2 -(n-1)/2
2^(n-1)*an =1/2
an= (1/2)^n
bn=n/an
= n(2)^n
= 2(n.2^(n-1))
consider
1+x+x^2+...+x^n = (x^(n+1)-1)/(x-1)
1+2x+..+nx^(n-1) = [(x^(n+1)-1)/(x-1)]'
= [nx^(n+1)-(n+1)x^n+1]/(x-1)^2
put x=1/2
1.(1/2)^0+2(1/2)^1+...+n(1/2)^(n-1)
=4[n(1/2)^(n+1)-(n+1)(1/2)^n+1]
=4( 1- (n+2)(1/2)^(n+1) )
Sn = b1+b2+..+bn
= summation (i:1->n) 2(i.2^(i-1))
= 8( 1- (n+2)(1/2)^(n+1) )