设bn=(2n-1)/(2^n),求数列{bn}的前n项和Tn.

bn=(2n-1)/(2^n),
Tn=1/2+3/2^2+5/2^3+7/2^4+……+(2n-3)/[2^（n-1)]+(2n-1)/(2^n),
2Tn=2/2+3/2^1+5/2^2+7/2^3+……+(2n-3)/[2^（n-2)]+(2n-1)/[2^(n-1)],
两式相减：
Tn=1+2/2^1+2/2^2+2/2^3+……+2/[2^（n-2)]+2/[2^(n-1)]-(2n-1)/(2^n)
=1+1*{1-1/[2^(n-1)]}/(1-1/2)-(2n-1)/(2^n)
=1+2{1-1/[2^(n-1)]}-(2n-1)/(2^n)
=3-2/[2^(n-1)]}-(2n-1)/(2^n)
=3-4/(2^n)-(2n-1)/(2^n)
=3-(2n+3)/(2^n)

这样的题目有个模式,bn=(2n-1)*1/(2^n) 也就是公比是1/2Tn=1/(2^1)+3/(2^2)+5/(2^3)+……+(2n-3)/(2^(n-1))+(2n-1)/(2^n) 一式通常遇到一个等差数列乘以一个等比数列,通用的方法是：先成公比；(1/2)*Tn=1/(2^2)+3/(2...