已知数列满足a1=1,根号an-1-根号an=根号anan-1,求an
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已知数列满足a1=1,根号an-1-根号an=根号anan-1,求an
已知数列满足a1=1,根号an-1-根号an=根号anan-1,求an
已知数列满足a1=1,根号an-1-根号an=根号anan-1,求an
√[a(n-1)]-√[an]=√[ana(n-1)]
两边同时除以√[ana(n-1)]
得:1/√[an]-1/√[an(n-1)]=1
令bn=1/√[an]
则bn-b(n-1)=1,b1=1
∴bn是以首相b1=1,公差d=1的等差数列
故bn=n
所以1/√[an]=n
得 an=1/n²
答案:an=1/n²
则bn-b(n-1)=1,b1=1
∴bn是以首相b1=1,公差d=1的等差数列
故bn=n
所以1/√[
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