等差数列{an},a1=1,前n项和Sn,S2n/Sn=4

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等差数列{an},a1=1,前n项和Sn,S2n/Sn=4
（1）求数列｛an｝的通项共识和Sn
（2）记bn=an*2^（n-1）求｛bn｝的前N项和Tn

S2n=2n+n*(2n-1)d
Sn=n+n(n-1)d/2
4Sn=4n+2(n^2-n)d
S2n/Sn=4 S2n=4Sn
4n+2d(n^2-n)=2n+(2n^2-n)d
整理,得
dn=2n
d=2
S2n=2n+n*(2n-1)d
Sn=n+n(n-1)d/2
4Sn=4n+2(n^2-n)d
S2n/Sn=4 S2n=4Sn
4n+2d(n^2-n)=2n+(2n^2-n)d
整理,得
dn=2n
d=2
an=1+(n-1)*2=2n-1
Sn=n(n+1)-n=n^2
bn=an*2^(n-1)=n^2*2^n/2

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