已知各项均为正数的数列an 前N项和为Sn 首项为a1，且2，an，Sn成等差数列

2a(n) = 2 + S(n),
2a(1) = 2 + S(1) = 2 + a(1), a(1) = 2.

2a(n+1) = 2 + S(n+1),
2a(n+1)-2a(n) = 2 + S(n+1) - 2 - S(n) = a(n+1),
a(n+1) = 2a(n),
{a(n)}是首项为a(1)=2,公比为2的等比数列。
a(n)=2*2^(n-1)=2^n, n = 1,2,...

b(n)=log(2)a(n)=n, n = 1,2,...
c(n)=b(n)/a(n) = n/2^n, n = 1,2,...

T(N) = c(1)+c(2)+...+c(N) = 1/2 + 2/2^2 + ... + (N-1)/2^(N-1) + N/2^N,
2T(N) = 1 + 2/2 + ... + (N-1)/2^(N-2) + N/2^(N-1),
T(N) = 2T(N) - T(N) = 1 + 1/2 + ... + 1/2^(N-1) - N/2^N = [1-1/2^N]/[1-1/2] - N/2^N = 2[1 - 1/2^N] - N/2^N = [2^(N+1)-2-N]/2^N

2+Sn=2an可得an=2^n
cn=log2(an)/an=n/2^n
Snc=2Snc-Snc=1+1-(1/2)^n-n/2^n