Since fibonacci numbers are a linear recurrence - and the initial conditions are special - we can express them by a matrix $$\begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}^n = \begin{pmatrix} F_{n+1} & F_n \\ F_n & F_{n-1} \end{pmatrix}$$ this is easy to prove by induction:
- $$\begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}^1 = \begin{pmatrix} F_{2} & F_1 \\ F_1 & F_{0} \end{pmatrix}$$
- $$\begin{pmatrix} F_{n+1} & F_n \\ F_n & F_{n-1} \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} F_n + F_{n+1} & F_{n+1} \\ F_n + F_{n-1} & F_{n} \end{pmatrix} = \begin{pmatrix} F_{n+2} & F_{n+1} \\ F_{n+1} & F_{n} \end{pmatrix}$$
Now your theorem follows from $$\begin{pmatrix} F_{n+1} & F_n \\ F_n & F_{n-1} \end{pmatrix}^2 = \begin{pmatrix} F_{n+1}^2 + F_n^2 & - \\ - & - \end{pmatrix} = \begin{pmatrix} F_{2n+1} & F_{2n} \\ F_{2n} & F_{2n-1} \end{pmatrix}$$
https://en.wikipedia.org/wiki/Fibonacci_number
Let $\phi=\dfrac{\sqrt{5}+1}2$ and note that $\phi^{-1} =\dfrac 1\phi= \dfrac{\sqrt{5}-1}2$.
Note also that $1+\dfrac 1\phi=\phi$ and $1-\phi=-\dfrac 1\phi$.
From your formula,
$$F_n = \frac 1{\sqrt{5}}\left[\phi^n-(-\frac 1\phi)^n \right]$$
For $n=k$ and $n=k-1$,
$$\begin{align}
F_k &= \frac 1{\sqrt{5}}\left[\phi^k-(-\frac 1\phi)^k \right]\\
F_{k-1} &= \frac 1{\sqrt{5}}\left[\phi^{k-1}-(-\frac 1\phi)^{k-1} \right]\\
&=\frac 1{\sqrt{5}} \left[\phi^k \cdot \frac 1\phi -(-\frac 1\phi)^k \cdot (-\phi)\right]\\
\end{align}$$
Hence,
$$\begin{align}
F_{k+1}&=F_{k}+F_{k-1}\\
&=\frac 1{\sqrt{5}} \left[\phi^k \cdot \left( 1+\frac 1\phi \right)
-(-\frac 1\phi)^k \cdot \left( 1-\phi \right)\right]\\
&=\frac 1{\sqrt{5}} \left[\phi^k \cdot \phi
-(-\frac 1\phi)^k \cdot \left( -\frac 1\phi \right)\right]\\
&=\frac 1{\sqrt{5}} \left[\phi^{k+1}-(-\frac 1\phi)^{k+1} \right]
\end{align}$$
i.e. if formula is true for $n=k-1$ and $n=k$, it is also true for $n=k+1$.
For $n=0$ and $n=1$, $F_0=0$ and $F_1=1$ respectively.
Hence $F_2=F_0+F_1=1$.
It can easily be shown that the formula is true for $n=2$.
Hence, by induction, formula is true for all positive integer $n\geq 2$.
Best Answer
$F_k=\frac{a^k-b^k}{\sqrt{5}}, a+b=1, ab=-1,a^2=a+1, b^2=b+1 $. These relations will be used frequently below. Then $$S_n=\sum_{j=1}^{n} F_{2n+1-j}~ {n \choose j}= \sum_{j=0}^{n} \frac{a^{2n+1-j}-b^{2n+1-j}}{\sqrt{5}} {n \choose j}.$$ Use binomial summation: $$\sum_{j=1}^{n} {n \choose k} a^{-j}=(1+1/a)^n-1$$ $$\implies S_n=\frac{1}{\sqrt{5}} [ a^{2n+1} ((1+1/a)^n-1)-b^{2n+1} ((1+1/b)^n-1)]$$ $$\implies S_n=\frac{1}{\sqrt{5}}\left [a^{2n+1}(1-b)^n-b^{2n+1}(1-a)^n]-[a^{2n+1}-b^{2n+1}]\right]$$ $$\implies S_n=\frac{1}{\sqrt{5}}[(a)^{3n+1}-(a)^{3n+1} ]-F_{2n+1}= F_{3n+1}-F_{2n+1}~~~(*)$$ We get a different result from that of OP. Our result can be checked numerically by hand or at Mathematica.