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an=Sn-Sn-1
因为:Sn=2^(n+1)-2,则
Sn-1=2^n-2
an=2^(n+1)-2^n=2^n
~~~~~~~~~~~~~~~~~
bn=(n+1)/2^n=n/2^n+1/2^n
对于前一项,令其前n项和为Kn
Kn=2/2+3/4+4/8+...+n/2^(n-1)+(n+1)/2^n,则
1/2Kn=2/4+3/8+...+n/2^n+(n+2)/2^(n+1)
两式相减
1/2Kn=2/2+(1/4+1/8+...+1/2^n)-(n+2)/2^(n+1)
=1+1/4*[(1-(1/2)^n]/(1-1/2)-(n+2)/2^(n+1)
=1+1/2-(1/2)^(n+1)-(n+2)/2^(n+1)
=3/2-(1/2)^(n+1)-(n+2)/2^(n+1)
则 Kn=3-(1/2)^n-(n+2)/2^n
后一项和为:
1/2*(1-(1/2)^n)/(1-1/2)=1-(1/2)^n
所以总和为:
4-n/2^n-2*(1/2)^(n-1)
因为:Sn=2^(n+1)-2,则
Sn-1=2^n-2
an=2^(n+1)-2^n=2^n
~~~~~~~~~~~~~~~~~
bn=(n+1)/2^n=n/2^n+1/2^n
对于前一项,令其前n项和为Kn
Kn=2/2+3/4+4/8+...+n/2^(n-1)+(n+1)/2^n,则
1/2Kn=2/4+3/8+...+n/2^n+(n+2)/2^(n+1)
两式相减
1/2Kn=2/2+(1/4+1/8+...+1/2^n)-(n+2)/2^(n+1)
=1+1/4*[(1-(1/2)^n]/(1-1/2)-(n+2)/2^(n+1)
=1+1/2-(1/2)^(n+1)-(n+2)/2^(n+1)
=3/2-(1/2)^(n+1)-(n+2)/2^(n+1)
则 Kn=3-(1/2)^n-(n+2)/2^n
后一项和为:
1/2*(1-(1/2)^n)/(1-1/2)=1-(1/2)^n
所以总和为:
4-n/2^n-2*(1/2)^(n-1)
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