记Cn=anbn求数列{Cn}的前n项和Tn

等差数列与等比数列相乘再求和
Tn=1×3/2^1+3×3/2^2+5×3/2^3+……+(2n-1)×3/2^n ①
两边同乘公比1/2（题目写的不太清楚,如果公比是3/2,方法是一样的）
(1/2)×Tn=(1/2)×1×3/2^1+3×(1/2)×3/2^2+(1/2)×5×3/2^3+……+(1/2)×(2n-1)×3/2^n
(1/2)×Tn=1×3/2^2+3×3/2^3+5×3/2^4+……+(2n-1)×3/2^(n+1) ②
①-②
(1/2)×Tn=3/2^1+[2×3/2^2+2×3/2^3+……+2×3/2^n]-(2n-1)×3/2^(n+1)
中括号内是等比数列
(1/2)×Tn=3/2^1+(3/2)[1-1/2^(n-1)]/(1-1/2)-(2n-1)×3/2^(n+1)
(1/2)×Tn=3/2+3-12/2^(n+1)]-(2n-1)×3/2^(n+1)
(1/2)×Tn=9/2-(2n+3)×3/2^(n+1)
Tn=9-3×(2n+3)/2^n