求和Sn=1*(1/3)+3*(1/3)^2+5*(1/3)^3+.+(2n-1)*(1/3)^n

等差乘等比求和的思想是乘积求差法:
数列乘公比再与原数列求差
Sn=1*(1/3)+3*(1/3)^2+5*(1/3)^3+.+(2n-1)*(1/3)^n
那么(1/3)*Sn=1*(1/3)^2+3*(1/3)^3+5*(1/3)^3+.+(2n-1)*(1/3)^(n+1)
上减下得
(2/3)*Sn=2*(1/3)^2+2*(1/3)^3+...+2*(1/3)^n+1*(1/3)-(2n-1)*(1/3)^(n+1)
除开最后两项则前面为等差数列:
求和得1-(1/3)^(n-1)
所以(2/3)*Sn=1-(1/3)^(n-1)+1*(1/3)-(2n-1)*(1/3)^(n+1)
把2/3乘到右边
再自己化简吧.

这是利用“错位相减消项”求数列的前n项和题型哦
第一步，两边乘以（1/3):
由Sn=1*(1/3) + 3*(1/3)^2 + 5*(1/3)^3 +....+(2n-1)*(1/3)^n ，得
(1/3)Sn=1*(1/3)^2 + 3*(1/3)^3 + 5*(1/3)^4 +....+[2(n-1)-1]*(1/3)^n+(2n-1)*(1/3)^(n+1) ...

an=5n-3
10Sn=an^2+5an+6
10S(n+1)=a(n+1)^2+5a(n+1)+6
两式相减得a(n+1)^2-an^2=5a(n+1)+5an
左右同除a(n+1)+an得
a(n+1)-an=5 这是个等差数列
a3=a1+10
a15=a1+70
又因为a3^2=a1*a15
即(a1+10)^2=a1*(a1+70)
解得a1=2
所以an=5n-3

Sn=1*(1/3) + 3*(1/3)^2 +....+(2n-1)*(1/3)^n (1)
Sn/3= 1*(1/3)^2 + ....+(2n-3)*(1/3)^n +(2n-1)*(1/3)^(n+1)(2)
(1)-(2),2Sn/3=1/3+2(1/9+1/27+……+（1/3)^n)-(2n-1)*(1/3)^(n+1)
2...