What is the summation notation for the Fibonacci numbers? I learned about summation notation the other day, and I'm looking for a way to write the Fibonacci numbers with it. What would it look like?
 A: $$F_n=\sum_{k=n-2}^{n-1}F_k$$
Given the initial conditions:
$$F_0=0$$
$$F_1=1$$
It's trivial, but it does use the summation notation.
A: The usual way to represent them is $F_{n+1}=F_n+F_{n-1}$, which doesn't involve $\sum$.  There are relations that involve it.  A couple of the ones given on the Wikipedia page are $$F_{n+2}=1+\sum_{i=1}^nF_i\\F_nF_{n+1}=\sum_{i=1}^nF_i^2$$
A: As noted, there isn't 'a' natural summation for the Fibonacci numbers (though Ataraxia's answer certainly comes closest to the traditional definition!), but there are many, many identities involving the Fibonaccis which can be written using summation notation.  My personal favorite — which is a good one to try and prove if you want practice in manipulating sums and with binomial numbers — is the formula $\displaystyle F_{n+1}=\sum_{i=0}^{n}{n-i \choose i}$.  Try circling the terms for $F_5$ and $F_6$, for instance, in Pascal's triangle, and see how they come together to give $F_7$; once you see it visually, that visual proof can be translated to a mathematical one.
A: By definition, we have $$F_0=0, F_1=1, \mbox{ and }F_{i+2}=F_{i+1}+F_i.$$
By taking the sum of the first $n$ Fibonacci numbers, we get $$\sum_{i=1}^n F_i=\sum_{i=1}^n (-F_{i+1}+F_{i+2})=$$ $$(-F_2+F_3)+(-F_3+F_4)+\cdots+(-F_{n}+F_{n+1})+(-F_{n+1}+F_{n+2})=$$ $$-F_2+(F_3-F_3)+(F_4-F_4)+\cdots+(F_{n+1}-F_{n+1})+F_{n+2}=F_{n+2}-F_2$$ and so we get $\sum_{i=1}^n F_i=F_{n+2}-1.$
Hence $$F_{n+2}=1+\sum_{i=1}^n F_i.$$
A: If you want to avoid the "..." notation
and get more exercise in summation notation,
Daniel Rust's post can be written like this:
$\begin{align}
\sum_{i=1}^n F_i
&=\sum_{i=1}^n (-F_{i+1}+F_{i+2})\\
&=\sum_{i=1}^n (-F_{i+1}) + \sum_{i=1}^n (F_{i+2})\\
&=-\sum_{i=2}^{n+1} F_{i} + \sum_{i=3}^{n+2} F_{i}\\
&=-(F_2+\sum_{i=3}^{n+1} F_{i}) + \sum_{i=3}^{n+1} F_{i}+F_{n+2}\\
&=F_{n+2}-F_2\\
\end{align}
$
Note that
the key is to separating out
the common part of the summations
($\sum_{i=3}^{n+1} F_{i}$)
so it can be cancelled out.
With practice,
this can be done
without all the explicit steps I have included.
A: $$\quad{\huge{1,1,2,3,5,8,13,21,34,55,...\\f_3=f_2+f_1\\f_4=f_3+f_2\\f_5=f_4+f_3\\f_6=f_5+f_4\\f_7=f_6+f_5\\\vdots\\f_{n}=f_{n-1}+f_{n-2}\\
\\f_{n+1}=f_{n}+f_{n-1}\\\\f_{n+2}=f_{n+1}+f_{n}\\f_{n+2}=f_2+\underbrace{f_1+f_2+f_3+f_4+...+f_n}_{S(n)}\\f_{n+2}=f_2+S(n)\\f_{n+2}=1+S(n)\\ \\s(n)=f_{n+2}-1}}$$
