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
 A: We know that $F_{n+2} = F_{n+1} + F_n \implies F_n = F_{n+2} - F_{n+1}$, so ...
$$\sum_{i=1}^nF_i = \sum_{i=1}^n\left(F_{i+2} - F_{i+1}\right)$$$$=\left(F_3 - F_2\right) + \left(F_4-F_3\right) + \dots + \left(F_{n+1} - F_{n}\right)+ \left(F_{n+2} - F_{n+1}\right)$$
$$= F_{n+2} + \left(F_{n+1} - F_{n+1}\right) + \dots + \left(F_{3} - F_{3}\right) - F_2$$

If you are determined to use induction, you have already provided a base case.. the step shouldn't be too crazy, you just need to massage some terms around  
Assume $\sum_{i=1}^nF_i = F_{n+2} - 1$, then
$$\sum_{i=1}^{n+1}F_i =F_{n+1} + \sum_{i=1}^nF_i = F_{n+1} + \left(F_{n+2} - 1\right)$$
$$= \left(F_{n+1} + F_{n+2}\right) - 1 = F_{n+3} - 1$$
A: $$
F_{3} = F_2 + F_{1} \\
F_{4} = F_3 + F_{2} \\
F_{5} = F_4 + F_{3}\\ 
...
F_{n} = F_{n -1} + F_{n-2}\\
F_{n+1} = F_n + F_{n-1}\\
F_{n+2} = F_{n+1} + F_{n}\\
$$
Now just sum the right side and the left side.
You'll get:
$$
 \sum_{i=3}^{n+2}F_i = \sum_{i=2}^{n+1}F_i + \sum_{i=1}^{n}F_i
$$
Then
$$
 \sum_{i=3}^{n}F_i + F_{n+1}+ F_{n+2} = \sum_{i=2}^{n+1}F_i + \sum_{i=3}^{n}F_i + F_1 + F_2
$$
A: For the base case you will need to use $n=1$ if you wish to prove for all $n\in\mathbb{Z^+}$. 
$$\sum_{i=1}^{n}{F_i} = \sum_{i=1}^{1}{F_i} = F_1 = 1$$
and 
$$F_{n+2}-1=F_3-1=2-1=1$$
Proving the base case.
For the induction step (weak induction suffices, note that weak induction is a special case of strong induction), you can assume the IH (induction hypothesis) for $n$ and prove the identity for $n+1$. Let $L(n)$ be the left-hand side of the identity for $n$ and let $R(n)$ be the right-hand side.
Then
$$\begin{align}
L(n+1) &= \sum_{i=1}^{n+1}{F_i} \\
       &= F_{n+1} + \sum_{i=1}^{n}{F_i} \\[1em]
       &= F_{n+1} + F_{n+2} - 1 &(\text{by IH for n}) \\[1em]
       &= F_{n+3} - 1 &(\text{by the recursion relationship}) \\[1em]
       &= R(n+1) 
\end{align}$$
This proves the induction step.
A: 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
$$F_{n+2}-1=\sum_{k=1}^n\,F_{n-k+1}=\sum_{i=1}^n\,F_i\,.$$
