a1=1,a(n+1)=2an,n=1,2,3,.求Sn
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a1=1,a(n+1)=2an,n=1,2,3,.求Sna1=1,a(n+1)=2an,n=1,2,3,.求Sna1=1,a(n+1)=2an,n=1,2,3,.求Sna(n+1)=2an所以an是等
a1=1,a(n+1)=2an,n=1,2,3,.求Sn
a1=1,a(n+1)=2an,n=1,2,3,.求Sn
a1=1,a(n+1)=2an,n=1,2,3,.求Sn
a(n+1)=2an
所以an是等比数列,q=2
a1=1
所以Sn=a1(1-q^n)/(1-q)=-1+2^n
∵a(n+1)=2an
∴﹛an﹜是等比数列
∴an=a1q^(n-1)=1×2^(n-1)=2^(n-1)
∴Sn=a1(1-q^n)/(1-q)=(1-2^n)/(1-2)=2^n-1
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