数列求和 an=（2n+1）/[2*3^（n-1）]

分析：错位相减
设前n项和为Sn,则Sn=3/[2*3^0]+5/[2*3^1]+。。。+(2n+1)/[2*3^(n-1)]
∴Sn/3=3/[2*3^1]+5/[2*3^2]+。。。+(2n+1)/[2*3^n]
∴Sn-Sn/3=3/[2*3^0]+2/[2*3^1]+。。。+2/[2*3^(n-1)]-(2n+1)/[2*3^n]
=2/3-(2n+1)/[2*3^n]+[1/(3^1)+1/(3^2)+。。。+1/(3^(n-1))]
=2/3-(2n+1)/[2*3^n]+1/3[1-(1/3)^(n-1)]/[1-1/3]
∴可以解得：Sn=3-(n+2)/[2*3^(n-1)]