a(n+1)=[f(√an)]^2,
a(n+1)=[√an/(√an+1)] ^2.
則√a(n+1)=√an/(√an+1)
1/√a(n+1)= (√an+1)/√an
即1/√a(n+1)=1+1/√an,
所以數(shù)列{1/√an }是等差數(shù)列,所以1/√an=1+(n-1)*1,
1/√an=n,an=1/n^2.
已知函數(shù)f(x)=x/(x+1),若數(shù)列{an}滿足：an>0,a1=1,an+1=[f(根號(hào)an)]^2?(1)求數(shù)列an的通項(xiàng)公式
已知函數(shù)f(x)=x/(x+1),若數(shù)列{an}滿足：an>0,a1=1,an+1=[f(根號(hào)an)]^2?(1)求數(shù)列an的通項(xiàng)公式
數(shù)學(xué)人氣：414 ℃時(shí)間：2020-04-17 19:02:13
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