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