1/a(n+1)=√1/an²+2
两边平方得
1/a(n+1)²=1/an²+2
则
1/an²是等差数列,公差d=2
则
1/an²=1/a1²+2(n-1)=2n-1
又an>0
则
an=1/√(2n-1)
已知数列﹛an﹜满足a1=1,1/a(n+1)=√1/an²+2,an>0,求an
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