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a function $f$ satisfies the following conditions:

$$\begin{align*} &f(1)=1\\ &f(n)=f(n-1)+2\sqrt{f(n-1)}+1\quad\text{for integers}\quad n\ge 2 \end{align*}$$

Find a formula that might be true for all integers $n\ge1$. Then prove using mathematical induction that it is indeed correct.

I found that the formula is $f(n)=n^2$, but I'm stuck on how to prove it... can anyone help me please?

Thanks a lot!

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It may or may not help to rewrite the recursion as $f(n)=(\sqrt{f(n-1)}+1)^2$ – Mike Oct 30 '12 at 0:57
ok thanks guys, I messed myself up with this recursion lol – user42624 Oct 30 '12 at 0:58
up vote 1 down vote accepted

HINT for the induction step: if $f(n)=n^2$, then $$f(n+1)=f(n)+2\sqrt{f(n)}-1=n^2+2\sqrt{n^2}+1=\dots~?$$

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Hint: you already have the base case. All you have to do is to use the induction hypothesis which is $f(n-1)=(n-1)^2$

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