cn=an/bn=2^n/[1/(2n-1)]=2^n×(2n-1)
Tn=2¹×1+2²×3+2³×5+·················+2^(n-1)×(2n-3)+2^n×(2n-1) ①
2Tn= 2²×1+2³×3+2^4×5+······+2^(n-1)×(2n-5)+2^n×(2n-3)+2^(n+1)×(2n-1) ②
②-①：Tn=-2-2×2²-2×2³-·············-2×2^(n-1)-2×2^n+2^(n+1)×(2n-1)
=-2×2²-2×2³-·············-2×2^(n-1)-2×2^n+2^(n+1)×(2n-1)-2
=-2×2¹-2×2²-2×2³-·············-2×2^(n-1)-2×2^n+2^(n+1)×(2n-1)-2+2×2¹
=-2×(2¹+2²+2³+···········+2^n)+2^(n+1)×(2n-1)-2+2×2¹
=-2×2×(1-2^n)/(1-2)+2^(n+1)×(2n-1)+2
=4(1-2^n)+2^(n+1)×(2n-1)+2
=4-2×2^(n+1)+2^(n+1)×(2n-1)+2
=2^(n+1)×(2n-3)+6
推广：类似等差乘以等比的数列求前N项和,只需乘以等比数列的公比,再错位相减,则得到
的新的等比数列和某几项特殊项.解题时注意书写,不然可能会减错.
祝愿楼主学业有成!