Prove $\sum^{k}_{i=0}{F(i)} + 1 = F(k+2)$ without induction I want to prove that $\sum^{k}_{i=0}{F(i)} + 1 = F(k+2)$, where $F(0) = 0$, $F(1) = 1$ and for all $n \geq 2$, $F(n) = F(n-1) + F(n-2)$ without using induction.
I want to prove it without induction because I want to implement the equation within another induction proof about Fibonacci numbers. Having 2 induction proofs together would be messy.

Use the general term formula: $$F_n=\frac{\phi^n-\psi^n}{\sqrt{5}}, \ \ \text{where} \ \ \phi=\frac{1+\sqrt{5}}{2}, \ \ \psi=\frac{1-\sqrt{5}}{2}.$$
Use the sum of geometric progression:
$$\begin{align}\sum^{n}_{i=0}{F(i)} + 1 &= \frac{\phi^0-\psi^0}{\sqrt{5}}+\frac{\phi^1-\psi^1}{\sqrt{5}}+\cdots + \frac{\phi^n-\psi^n}{\sqrt{5}}+1=\\
&=\frac{1}{\sqrt{5}}([\phi^1+\phi^2+\cdots +\phi^n]-[\psi^1+\psi^2+\cdots +\psi^n])+1=\\
&=\frac{1}{\sqrt{5}}\cdot \left(\frac{\phi(\phi^n-1)}{\phi-1}-\frac{\psi(\psi^n-1)}{\psi-1}\right)+1=\\
&=\frac{1}{\sqrt{5}}\cdot \left(\frac{\phi(\phi^n-1)}{-\psi}-\frac{\psi(\psi^n-1)}{-\phi}\right)+1=\\
&=\frac{1}{\sqrt{5}}\cdot \frac{\phi^2(\phi^n-1)-\psi^2(\psi^n-1)}{-\psi\cdot \phi}+1=\\
&=\frac{\phi^{n+2}-\psi^{n+2}}{\sqrt{5}}+ \frac{-\phi^2+\psi^2}{\sqrt{5}}+1=\\
&=\frac{\phi^{n+2}-\psi^{n+2}}{\sqrt{5}}=\\
&=F(n+2).\end{align}$$
