高一数学题，第九题
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an=2n-1 , bn=2^n
an/bn=(2n-1)/2^n
Sn=1/2+3*(1/2)^2+5*(1/2)^3+...+(2n-1)*(1/2)^n
1/2Sn=(1/2)^2+3*(1/2)^3+...+(n-1)*(1/2)^n+(2n-1)(1/2)^(n+1)
两式相减得
1/2Sn=1/2+2[(1/2)^2+(1/2)^3+...+(1/2)^n]-(2n-1)(1/2)^(n+1)
1/2Sn=1/2+2[1/2(1-(1/2)^n]-(2n-1)(1/2)^(n+1)
1/2Sn=1/2+(1-(1/2)^n-(2n-1)(1/2)^(n+1)
1/2Sn=3/2-(1/2)^n-(2n-1)(1/2)^(n+1)
1/2Sn=3/2-(1/2)^n-(2n-1)/2*(1/2)^n
1/2Sn=3/2-(n-1)*(1/2)^n
Sn=3-[(n-1)*(1/2)^(n-1)]
an/bn=(2n-1)/2^n
Sn=1/2+3*(1/2)^2+5*(1/2)^3+...+(2n-1)*(1/2)^n
1/2Sn=(1/2)^2+3*(1/2)^3+...+(n-1)*(1/2)^n+(2n-1)(1/2)^(n+1)
两式相减得
1/2Sn=1/2+2[(1/2)^2+(1/2)^3+...+(1/2)^n]-(2n-1)(1/2)^(n+1)
1/2Sn=1/2+2[1/2(1-(1/2)^n]-(2n-1)(1/2)^(n+1)
1/2Sn=1/2+(1-(1/2)^n-(2n-1)(1/2)^(n+1)
1/2Sn=3/2-(1/2)^n-(2n-1)(1/2)^(n+1)
1/2Sn=3/2-(1/2)^n-(2n-1)/2*(1/2)^n
1/2Sn=3/2-(n-1)*(1/2)^n
Sn=3-[(n-1)*(1/2)^(n-1)]
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