Question

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

Answer 1

两边同乘3
3Sn=1+3*(1/3)+5*(1/3)^2+...+(2n-1)*(1/3)^(n-1)
Sn = 1*(1/3)+3*(1/3)^2+...+(2n-3)*(1/3)^(n-1)+(2n-1)*(1/3)^n
两式相减
2Sn=1+2*(1/3)+2*(1/3)^2+...+2*(1/3)^(n-1)-(2n-1)*(1/3)^n
2Sn=1-(2n-1)*(1/3)^n+2*[1/3+(1/3)^2+...+(1/3)^(n-1)]
方括号内是一等比数列求和，首项1/3,公比1/3,一共(n-1)项（看指数1,2,...,n-1）
2Sn=1-(2n-1)*(1/3)^n+2*(1/3)(1-(1/3)^(n-1))/(1-1/3)
2Sn=1-(2n-1)*(1/3)^n+1-(1/3)^(n-1)
=2-(2n-1+3)*(1/3)^(n)
=2-(2n+2)*(1/3)^(n)
Sn=1-(n+1)*(1/3)^(n)

Answer 2

sn=1*1/3+3*(1/3)^2+5*（1/3）^3+......+(2n-1)*(1/3）^n
sn/3=1*1/3^2+3*(1/3)^3+5*（1/3）^4+......+(2n-1)*(1/3）^(n+1)
sn-sn/3=1/3+2/3^2+2/3^3+...........+2/3^n-(2n-1)*(1/3）^(n+1)
2sn/3=-1/3+2/3^1+2/3^2+2/3^3+...........+2/3^n-(2n-1)*(1/3）^(n+1)
2sn/3=-1/3+2/3*[1-(1/3)^n]/(1-1/3)-(2n-1)*(1/3）^(n+1)
2sn/3=-1/3+1-(1/3)^n-(2n-1)*(1/3）^(n+1)
2sn/3=2/3-(1/3)^n-(2n-1)/3^(n+1)
2sn/3=2/3-3*1/3^(n+1)-(2n-1)/3^(n+1)
2sn/3=2/3-(2n+2)/3^(n+1)
sn/3=1/3-(n+1)/3^(n+1)
sn=1-(n+1)/3^n

Answer 3

把Sn乘以1/3，然后用错位相减法就OK