已知数列cn 设an=2n-3 ,bn=2的n-3次方 cn=an/bn,求数列cn前n项和

c(n) = (2n-3)/2^(n-3),
t(n) = c(1) + c(2) + c(3) + ... + c(n-1) + c(n)
= (2*1-3)/2^(1-3) + (2*2-3)/2^(2-3) + (2*3-3)/2^(3-3) + ... + [2(n-1)-3]/2^(n-4) + (2n-3)/2^(n-3),
2t(n) = (2*1-3)/2^(0-3) + (2*2-3)/2^(1-3) + (2*3-3)/2^(2-3) + ... + [2(n-1)-3]/2^(n-5) + (2n-3)/2^(n-4),
t(n) = 2t(n) - t(n) = (2*1-3)/2^(0-3) + 2/2^(1-3) + 2/2^(2-3) + ... + 2/2^(n-4) - (2n-3)/2^(n-3)
= 2/2^(0-3) + 2/2^(1-3) + ... + 2/2^(n-1-3) - 3/2^(-3) - (2n-3)/2^(n-3)
= 16[1 + 1/2 + ... + 1/2^(n-1)] - 24 - (2n-3)/2^(n-3)
= 16[1 - 1/2^n]/(1-1/2) - 24 - (2n-3)/2^(n-3)
= 32[1- 1/2^n] - 24 - (2n-3)/2^(n-3)
= 8 - 4/2^(n-3) - (2n-3)/2^(n-3)
= 8 - (2n+1)/2^(n-3)