b(1)=a(2^1)=a(2)=5
b(2)=a(2^2)=a(4)=9
b(3)=a(2^3)=a(8)=17
b(4)=a(2^4)=a(16)=33
...
...
b(n)=a(2^n)=2*(2^n)+1= 2^(n+1)+1
于是很显然b(n)的前n项和：
Tn=b(1)+b(2)+...+b(n)
=[2^2+1]+[2^3+1]+[2^4+1]+...+[2^(n+1)+1]
=[2^2+2^3+2^4+2^(n+1)] + n
(前面n项为等比数列)
=[(2^2)(1-2^n)/(1-2)] +n
=2^(n+2) - 4 + n