Consider the Fibonacci sequence $\{a_{n}\}$ Use mathematical induction to prove that $a_{n+1}a_{n-1}=(a_{n})^{2}+(-1)^{n}$
So far, I have tested the base case $n=1$ which is true. I am stuck on the inductive step where I plug in $k=n+1$.
$a_{n+2}a_{n}=(a_{n+1})^{2}+(-1)^{n+1}$
$a_{n+2}a_{n}=(a_{n+1})^{2}-(-1)^{n}$
I am unsure what the next step to take is.
Recall that: $a_1=1$,$a_2=1$,$a_3=2$ .
For $n=2$: $$ a_3a_1=a_2^2+(-1)^2=2 $$
We assume the hypothesis valid for $n=k$, such that: $$ a_{k+1}a_{k-1}=a_k^2+(-1)^k $$
Let's find the value of $a_{k+2}a_{k}$. We can use the Fibonacci recursion: $a_{k+2}=a_{k+1}+a_k$. Therefore: $$ a_{k+2}a_{k}=(a_{k+1}+a_k)a_k=a_{k+1}a_{k}+a_k^2 $$ From the hypothesis: $$ a_{k+1}a_{k-1}=a_k^2+(-1)^k \Rightarrow a_k^2 = a_{k+1}a_{k-1}-(-1)^k $$
Thus: $$ a_{k+2}a_{k}=(a_{k+1}+a_k)a_k=a_{k+1}a_{k}+a_{k+1}a_{k-1}-(-1)^k $$ $$ a_{k+2}a_{k}=a_{k+1}a_{k}+a_{k+1}a_{k-1}+(-1)^{k+1}=a_{k+1}(a_{k}+a_{k-1})+(-1)^{k+1} $$ Apply Fibonacci recursion again to find: $$ a_{k+2}a_{k}=a_{k+1}^2+(-1)^{k+1} $$