Question

数列，求解

Answer 1

(1)
f(x)= x/(2x+1)
a1=1
a(n+1)=f(an)
= an/(2an +1)
1/a(n+1) = (2an +1)/an
= 2 + 1/an
1/a(n+1) -1/an =2
{1/an }是等差数列, d=2
1/an -1/a1 =2(n-1)
1/an = 2n-1
an = 1/(2n-1)
(2)
let
S = 1.2^1+2.2^2+...+n.2^n (1)
2S = 1.2^2+2.2^3+...+n.2^(n+1) (2)
(2)-(1)
S = n.2^(n+1) -(2+2^2+...+2^n)
=n.2^(n+1) -2(2^n-1)

an = 2^n/an
= 2^n .(2n-1)
=2(n.2^n) - 2^n
Sn = a1+a2+...+an
=2S - 2(2^n-1)
=2n.2^(n+1) -6(2^n-1)
= 6 + (4n-6).2^n