数列{an}中,a1=1,根号an-根号a(n+1)=根号【ana(n+1)】,则{an}的通项an

问题描述:

数列{an}中,a1=1,根号an-根号a(n+1)=根号【ana(n+1)】,则{an}的通项an

√an-√a(n+1)=√[ana(n+1)]
1/√a(n+1)-1/√an=1
1/√a(n+1)=1+1/√an
1/√an
=1/√a(n-1)+1
=1/√a(n-2)+2
=...
=1/√a1+n-1
=n
an=1/n²
如仍有疑惑,欢迎追问.祝:。。。没看懂啊√an-√a(n+1)=√[ana(n+1)]1/√a(n+1)-1/√an=1 (把等号右边的东西除到左边去)1/√a(n+1)=1+1/√an (移项。。)1/√an(递推。。)=1/√a(n-1)+1=1/√a(n-2)+2=...=1/√a1+n-1=nan=1/n² (完事)