已知{an}满足an+a(n+1)=2a(n+2,)且a1=1,a2=2,设bn=a(n+1)-an,证明{bn}是等比数列
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已知{an}满足an+a(n+1)=2a(n+2,)且a1=1,a2=2,设bn=a(n+1)-an,证明{bn}是等比数列已知{an}满足an+a(n+1)=2a(n+2,)且a1=1,a2=2,设
已知{an}满足an+a(n+1)=2a(n+2,)且a1=1,a2=2,设bn=a(n+1)-an,证明{bn}是等比数列
已知{an}满足an+a(n+1)=2a(n+2,)且a1=1,a2=2,设bn=a(n+1)-an,证明{bn}是等比数列
已知{an}满足an+a(n+1)=2a(n+2,)且a1=1,a2=2,设bn=a(n+1)-an,证明{bn}是等比数列
证明 令bn=a(n+1)-an
2a(n+2)=an+a(n+1)
∴2[a(n+2)-a(n+1)]=an-a(n+1)=-[a(n+1)-an]
bn=a(n+1)-an,∴2b(n+1)=-bn,即b(n+1)/bn=-1/2
∴{bn}是等比数列
因为,
An+An+1=2An+2
故 ( An+An+1)/2-An+1=An+2-An+1
即 Bn+1=-1/2(An+1-An)=-1/2 Bn
{Bn}为等比数列,公比为-1/2 ,首项为1
a(n+2)-an=1/2[a(n+1)-an]
a(n+2)-a(n+1)+a(n+1)-an=1/2[a(n+1)-an]
b(n+1)+bn=1/2bn
b(n+1)=-1/2bn
所以bn=a(n+1)-a(n)是的以b1=a2-a1=1为首项,公比为-1/2的等比数列
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