Question

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

Answer 1

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.