已知等差数列{An},前n项和为Sn.A3=6,S3=12.求数列{2^(n-1)An}的前n项和Bn.已知等差数列{An},前n项和为Sn.A3=6,S3=12.求数列{2^(n-1)An}的前n项和Bn.

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$已知等差数列{An},前n项和为Sn.A3=6,S3=12.求数列{2^(n-1)An}的前n项和Bn.已知等差数列{An},前n项和为Sn.A3=6,S3=12.求数列{2^(n-1)An}的前n项和Bn.$

已知等差数列{An},前n项和为Sn.A3=6,S3=12.求数列{2^(n-1)An}的前n项和Bn.已知等差数列{An},前n项和为Sn.A3=6,S3=12.求数列{2^(n-1)An}的前n项和Bn.
已知等差数列{An},前n项和为Sn.A3=6,S3=12.求数列{2^(n-1)An}的前n项和Bn.
已知等差数列{An},前n项和为Sn.
A3=6,S3=12.
求数列{2^(n-1)An}的前n项和Bn.

已知等差数列{An},前n项和为Sn.A3=6,S3=12.求数列{2^(n-1)An}的前n项和Bn.已知等差数列{An},前n项和为Sn.A3=6,S3=12.求数列{2^(n-1)An}的前n项和Bn.
S3=a1+a2+a3=3a2=12
a2=4
d=a3-a2=6-4=2
a1=a2-d=4-2=2
an=a1+(n-1)d=2+2(n-1)=2n
令bn=2^(n-1)an,则bn=2n×2^(n-1)=n×2ⁿ
Bn=b1+b2+...+bn=1×2+2×2²+3×2³+...+n×2ⁿ
2Bn=1×2²+2×2³+...+(n-1)×2ⁿ +n×2^(n+1)
Bn-2Bn=-Bn=2+2²+...+2ⁿ-n×2^(n+1)
Bn=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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