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）

来源：学生作业帮编辑：大师作文网作业帮分类：数学作业时间：2024/09/26 00:26:13

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

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