Fibonacci proof question $\sum_{i=1}^nF_i = F_{n+2} - 1$ The sequence of numbers $F_n$ for $n \in N$ defined below are called the Fibonnaci numbers.
 $F_1 = F_2 = 1$, and for $n \geq 2$, $F_{n+1} = F_n + F_{n-1}$.
 Prove the following facts about the Fibonnaci numbers
For each of the following, $n \in N$
a) $\displaystyle \sum_{i=1}^nF_i = F_{n+2} - 1$
I started this proof by using strong induction
To prove something by strong induction, you have to prove that if all natural numbers strictly less than $N$ have the property, then $N$ has the property.
$n \geq 2$, $F_{n+1} = F_n + F_{n-1}$
Check basis step $n=2$: 
$F_1 + F_2 = 1 + 1 = 2 = 3-1 = F_4 - 1$, therefore TRUE
I'm unsure how to go any further, when it comes to strong induction I'm lost on how to set up my IH, and proceed for the rest of the steps

Combinatorial Proof: 
For each $n\in\mathbb{N}_0$, the number of ways to tile an $1$-by-$n$ array with $1$-by-$1$ squares and $1$-by-$2$ dominos is $F_{n+1}$.  Hence, the number of ways to tile an $1$-by-$(n+1)$ array in such a manner with at least one domino is given by $F_{n+2}-1$.
Now, let $k$ be the earliest position of the dominos.  For each $k$, we are left with the $1$-by-$(n-k)$ array to tile, which can be done in $F_{n-k+1}$ ways.  Since $k$ can be anything from $1$ to $n$, we have
$$F_{n+2}-1=\sum_{k=1}^n\,F_{n-k+1}=\sum_{i=1}^n\,F_i\,.$$
