Question

求高手帮忙解下数列题

Answer 1

(1)
Sn=2an-2^(n+1)+2
n=1,a1= 2^2-2=2

an=Sn-S(n-1)
=2an -2^(n+1) -2a(n-1) +2^n
an = 2a(n-1)+2^n
an/2^n -a(n-1)/2^(n-1) =1
{an/2^n} 是等差数列, d=1
an/2^n-a1/2^1=n-1
an/2^n =n
an = n.2^n
(2)
bn=log<2>(a1/1) + log<2>(a2/2) +...+log<2>(an/n)
=1+2+...+n
=n(n+1)/2
cn =1/bn
= 2/[n(n+1)]
= 2[1/n -1/(n+1)]
Tn =c1+c2+...+cn
= 2( 1/1- 1/(n+1)]
= 2n/(n+1)