Induction proof of $F(n)^2+F(n+1)^2=F(2n+1)$, where $F(n)$ is the $n$th Fibonacci number. 
Let $F(n)$ denotes the $n$th number in Fibonacci sequence.  Then for all $n\in\mathbb{N}$,
  $$F(n)^2+F(n+1)^2=F(2n+1).$$

I know how to prove it by using the formula 
$$F(n)=\frac{\left(\frac{1+\sqrt5}{2}\right)^n-\left(\frac{1-\sqrt5}{2}\right)^n}{\sqrt{5}},$$
but is there a way to prove it by induction?
I am year 12 standard so please don't go too deep.

The base cases $n=0$ and $n=1$ are easy to verify.
Note that
\begin{align*}
F(2n+3)&=F(2n+2)+F(2n+1)=(F(2n)+F(2n+1))+F(2n+1)\\
&=F(2n)+2F(2n+1)\\
&=(F(2n+1)-F(2n-1))+2F(2n+1)\\&=
3F(2n+1)-F(2n-1).
\end{align*}
Hence, by the inductive hypothesis,
\begin{align*}
F(2n+3)&=3F(2n+1)-F(2n-1)=3(F(n)^2+F(n+1)^2)-(F(n-1)^2+F(n)^2)\\
&=2F(n)^2+3F(n+1)^2-F(n-1)^2\stackrel{?}{=}F(n+1)^2+F(n+2)^2.
\end{align*}
So it remains to show that
$$F(n+2)^2+F(n-1)^2\stackrel{?}{=}2F(n)^2+2F(n+1)^2$$
which holds because
\begin{align*}
F(n+2)^2+F(n-1)^2&=(F(n+1)+F(n))^2+(F(n+1)-F(n))^2\\&=2F(n)^2+2F(n+1)^2.
\end{align*}
