已知等差数列{an}的前n项和为Sn,a2=1,S11=33.设bn=an/2^n,求数列{bn}的前n项和Tn

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$已知等差数列{an}的前n项和为Sn,a2=1,S11=33.设bn=an/2^n,求数列{bn}的前n项和Tn$

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已知等差数列{an}的前n项和为Sn,a2=1,S11=33.设bn=an/2^n,求数列{bn}的前n项和Tn
已知等差数列{an}的前n项和为Sn,a2=1,S11=33.
设bn=an/2^n,求数列{bn}的前n项和Tn

已知等差数列{an}的前n项和为Sn,a2=1,S11=33.设bn=an/2^n,求数列{bn}的前n项和Tn
因为S11=33,所以a6=3
又因为a2=1
所以an=n/2
所以bn=n/2^(n+1)
所以Tn=1/2^2+…+n/2^(n+1)
所以2Tn=1/2+2/2^2+…+n/2^n
两式相减得Tn=1/2+1/2^2+…+1/2^n-n/2^(n+1)
=1-(n-2)/2^(n+1)

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