Question

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

Answer 1

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) )