已知数列{an}满足,a1=0,a(n+1)=an+1+2根号下(an+1),则an=.（求通项）

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已知数列{an}满足,a1=0,a(n+1)=an+1+2根号下(an+1),则an=.（求通项）
已知数列{an}满足,a1=0,a(n+1)=an+1+2根号下(an+1),则an=.（求通项）

已知数列{an}满足,a1=0,a(n+1)=an+1+2根号下(an+1),则an=.（求通项）
a(n+1)=an+1+2√(an+1)
a(n+1)+1=(√(an+1)+1)^2
{√[a(n+1)+1]}^2=(√(an+1)+1)^2
{√[a(n+1)+1+√(an+1)+1)}{√[a(n+1)+1-√(an+1)-1)}=0
√(n+1)+1+√(an+1)+1)=0 或 √[a(n+1)+1-√(an+1)-1]=0
√[a(n+1)+1-√[(an+1]-1]=0
√[a(n+1)+1-√[(an+1=1
{√(an+1}成等差
√(an+1)=√(a1+1)+(n-1)=1+n-1=n
an=n^2-1

由a(n+1)=an+1+2根号下(an+1）得，a（n+1）+1=[根号下(an+1）+1]²。
而a（n+1）+1可以写成根号下[a(n+1)+1]²，所以根号下a(n+1)+1=根号下(an+1）+1
也就是说根号下（an+1）成等差数列。
根号下（a1+1）=1，所以根号下（an+1）=n，
∴an=n²-1

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