已知数列{An}是等差数列前n项和Sn,A3=6,S3=12.求数列{An}的通项公式.求数列{1/Sn}的前n项和Bn.求数列{(2的n-1次方)An}的前n项和Cn.

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已知数列{An}是等差数列前n项和Sn,A3=6,S3=12.求数列{An}的通项公式.求数列{1/Sn}的前n项和Bn.求数列{(2的n-1次方)An}的前n项和Cn.
已知数列{An}是等差数列前n项和Sn,A3=6,S3=12.
求数列{An}的通项公式.
求数列{1/Sn}的前n项和Bn.
求数列{(2的n-1次方)An}的前n项和Cn.

已知数列{An}是等差数列前n项和Sn,A3=6,S3=12.求数列{An}的通项公式.求数列{1/Sn}的前n项和Bn.求数列{(2的n-1次方)An}的前n项和Cn.
1、a3=6,S3=12
所以a1+2d=6,3a1+3d=12
所以a1=2,d=2
所以{an}的通项公式为an=2+(n-1)*2=2n
2、Sn=2n+n(n-1)=n²+n=n(n+1)
1/Sn=1/n-1/(n+1)
Bn=1/S1+1/S2+······+1/Sn
=1-1/2+1/2-1/3+1/3-1/4+······+1/n-1/(n+1)
=1-1/(n+1)
=n/(n+1)
3、令Cn=2^(n-1)an,则Cn=2n×2^(n-1)=n×2ⁿ
Cn=C1+C2+...+Cn=1×2+2×2²+3×2³+...+n×2ⁿ
2Cn=1×2²+2×2³+...+(n-1)×2ⁿ +n×2^(n+1)
Cn-2Cn=-Cn=2+2²+...+2ⁿ-n×2^(n+1)
Cn=n×2^(n+1)-(2+2²+...+2ⁿ)
=n×2^(n+1)-2×(2ⁿ-1)/(2-1)
=n×2^(n+1)-2^(n+1) +2
=(n-1)×2^(n+1) +2

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