<s>[INST] For the Fibonacci sequence prove that $\sum_{i=1}^n F_i= F_{n+2} - 1$ 
For the Fibonacci sequence $F_1=F_2=1$, $F_{n+2}=F_n+F_{n+1}$,  prove that $$\sum_{i=1}^n  F_i= F_{n+2} - 1$$ for $n\ge 1$.

I don't quite know how to approach this problem. 
Can someone help and explain please?
 [/INST] Just for fun, let us try to prove $\sum\limits_{k=1}^n  F_k= F_{n+2} - 1$ using Binet's formula.

Recall that we have
$$F_n=\frac{a^n-b^n}{\sqrt5}$$
where $a=\frac{1+\sqrt5}2$ and $b=\frac{1-\sqrt5}2$.
A few identities, that are easy to check and which may be useful:
$$
\begin{align*}
a+b&=1\\
a-b&=\sqrt5\\
ab&=-1\\
1+a&=a^2\\
1+b&=b^2
\end{align*}
$$
(Most of these identities simply follow from noticing that $a$, $b$ are roots of $x^2-x-1$ and using Vieta's formulas).

Now we want to compute
$$\sum\limits_{k=1}^n  F_k=
\sum\limits_{k=1}^n  \frac{a^k-b^k}{\sqrt5}=
\frac 1{\sqrt5} \left(\sum\limits_{k=1}^n  a^k- \sum\limits_{k=1}^n  b^k\right)
$$
Using the formula $1+x+\dots+x^n = \frac{x^{n+1}-1}{x-1}$ we get
$$
\begin{gather*}
\frac1{\sqrt5} \left(\frac{a^{n+1}-1}{a-1}-\frac{b^{n+1}-1}{b-1}\right) = \\
\frac1{\sqrt5} \cdot \frac{(a^{n+1}-1)(b-1)-(b^{n+1}-1)(a-1)}{(a-1)(b-1)} = \\
\frac1{\sqrt5} \cdot \frac{a^{n+1}b-a^{n+1}-b-b^{n+1}a+b^{n+1}+a}{ab-a-b+1} = \\
\frac1{\sqrt5} \cdot \frac{-(a^{n+1}-b^{n+1})+ab(a^n-b^n)+(a-b)}{ab-(a+b)+1} = \\
\frac1{\sqrt5} \cdot \frac{-(a^{n+1}-b^{n+1})-(a^n-b^n)+\sqrt5}{-1-1+1} = \\
\frac{-(a^{n+1}-b^{n+1})-(a^n-b^n)+\sqrt5}{-\sqrt5}= \\
\frac{(a^{n+1}-b^{n+1})+(a^n-b^n)-\sqrt5}{\sqrt5}= \\
\frac{a^{n+1}-b^{n+1}}{\sqrt5}+\frac{a^{n}-b^{n}}{\sqrt5}-\frac{\sqrt5}{\sqrt5} = \\ 
F_{n+1}+F_n-1 = F_{n+2}-1
\end{gather*}
$$
</s>