已知an=2n-1,bn=2^n,记Cn=anbn,求数列Cn的前n项和?

记S=∑Cn=1×2+3×2^2+5×2^3+.+(2n-1)×2^n
那么2S=1×2^2+3×2^2+5×2^4+.+(2n-1)×2^(n+1)
两式相减得 S=1×2-2×2^2-2×2^3-.-2×2^n +(2n-1)2^(n+1)
= 2 -2^3[1+2+.+2^(n-2)]+(2n-1)2^(n+1)
=2-2^3[2^(n-1)-1]+(2n-1)2^(n+1)
=(2n-1)2^(n+1)-2^(n+2)-6
=(2n-3)2^(n+1)-6