已知数列an满足a1=1/2,anan+1=2an-an+1,求an
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已知数列an满足a1=1/2,anan+1=2an-an+1,求an
n+1是下角标
n+1是下角标
an*a(n+1)=2an-a(n+1)
--->1+1/an=2/a(n+1)
设:bn=1/an,则2b(n+1)=bn+1
2[b(n+1)-1]=bn-1
[b(n+1)-1]/[bn-1]=1/2
由此可看出{bn-1}为公比为1/2的等比数列
那么bn-1=(b1-1)*(1/2)^(n-1) =(1/a1-1)*(1/2)^(n-1) =-(1/2)^n
则bn=1-(1/2)^n,故an=1/[1-(1/2)^n]=[2^n]/[2^n-1]
--->1+1/an=2/a(n+1)
设:bn=1/an,则2b(n+1)=bn+1
2[b(n+1)-1]=bn-1
[b(n+1)-1]/[bn-1]=1/2
由此可看出{bn-1}为公比为1/2的等比数列
那么bn-1=(b1-1)*(1/2)^(n-1) =(1/a1-1)*(1/2)^(n-1) =-(1/2)^n
则bn=1-(1/2)^n,故an=1/[1-(1/2)^n]=[2^n]/[2^n-1]
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