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I would appreciate if somebody could help me with the following problem:

Question: Defined by $a_{1} =1,a_{2}=2$ and $a_n a_{n+2}=a_{n+1}^2+1(n\geq 1)$

Proof. $\frac{a_n^2+a_{n+1}^2+1}{a_{n}a_{n+1}} $ is constant for $n\geq 1$

I tried (Mathematical Induction) but couldn’t get it that way.

Young
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2 Answers2

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$b_n:=\dfrac{a_n^2+(a_{n+1}^2+1)}{a_{n}a_{n+1}}=\dfrac{a_n^2+(a_{n}a_{n+2})}{a_{n}a_{n+1}}=\dfrac{a_n+a_{n+2}}{a_{n+1}}$

and

$b_{n+1}=\dfrac{(a_{n+1}^2)+a_{n+2}^2+(1)}{a_{n+1}a_{n+2}}=\dfrac{(a_na_{n+2})+a_{n+2}^2}{a_{n+1}a_{n+2}}=\dfrac{a_n+a_{n+2}}{a_{n+1}}=b_n$

Nikolaj-K
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The formula $$\frac{a_n^2+a^2_{n+1}+1}{a_na_{n+2}}=3$$ holds for $n=1,2$. Assume it is true for $k$ and show it is true for $k+1$. Here is another approach:

$$ \frac{a_{k+1}^2+a^2_{k+2}+1}{a_{k+1}a_{k+2}}\cdot\frac{a_{k}}{a_{k}} =\frac{a_{k+1}^2+a^2_{k+2}+1}{a_{k}a_{k+2}}\cdot\frac{a_{k}}{a_{k+1}} =\frac{a_{k+1}^2+a^2_{k+2}+1}{a_{k+1}^2+1}\cdot\frac{a_{k}}{a_{k+1}} $$ $$ =\frac{a_k}{a_{k+1}}\left(\frac{a_{k+1}^2+1}{a_{k+1}^2+1}+\frac{a_{k+2}^2}{a_{k+1}^2+1}\right) =\frac{a_k}{a_{k+1}}\left(1+\frac{a_{k+2}^2}{a_k a_{k+2}}\right) =\frac{a_k+a_{k+2}}{a_{k+1}}. $$

So now we need to show that is constant in $k$. See solution above by NikolajK.

jdods
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