Sn=2an+2n,a1=-1,求Sn
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Sn=2an+2n,a1=-1,求Sn
Sn=2an+2n,a1=-1,求Sn
Sn=2an+2n,a1=-1,求Sn
S(n-1)=2a(n-1)+2(n-1),an=Sn-S(n-1)=2an-2a(n-1)+2,那么移项
an=2a(n-1)-2,左右两边添-2
an-2=2{a(n-1)-2},那么an-2是一个公比为2的等比数列,接下来,一共n项,就是Sn-2n=(-3)(2^n-1),-3是首项,再做整理可得出答案
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