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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