Summation of a constant using sigma notation Apologies if this is a silly question, but is it possible to prove that $$\sum_{n=1}^{N}c=N\cdot c$$ or does this simply follow from the definition of sigma notation? 
I am fairly sure it's the latter, but for some reason I've managed to get myself thrown by the absence of a summation index (intuitively of course it makes sense that summing a constant $N$-times should equal $N\cdot c$).
 A: It is possible to prove it. If you define $\sum$ notation recursively then it's something you can prove by induction. Specifically, given a sequence $a_1, a_2, \dots$ of numbers, you can define $\displaystyle\sum_{i=1}^n a_i$ recursively by:
$$\displaystyle\sum_{i=1}^1 a_i = a_1 \quad \text{and} \quad \displaystyle\sum_{i=1}^{n+1} a_i = \left( \sum_{i=1}^n a_i \right) + a_{n+1} \text{ for all } n \in \mathbb{N}$$
An inductive proof that $\displaystyle\sum_{i=1}^n c = nc$ then proceeds as follows:


*

*(Base case) $\displaystyle\sum_{i=1}^1 c = c$;

*(Induction step) Fix $n$ and suppose $\displaystyle\sum_{i=1}^n c = nc$. Then
$$\sum_{i=1}^{n+1} c = \left( \sum_{i=1}^n c \right) + c = nc + c = (n+1)c$$


So we're done. Here, the sequence was just the constant sequence $c,c,c,\dots$.
A: You can write it as $c\sum_{i=1}^N 1$ and then say that the sum of $N$ $1's$ is $N$, but that doesn't actually shed much light on the matter.
The thing to remember, ultimately, is that $\sum$ isn't some abstruse function. It's just a more compact way of writing addition. Writing $\sum_{x=1}^N f(x)$ means writing $f(1)+f(2)+\cdots f(N)$, and anything that you can do with addition you can do with it, because it is exactly addition.
A: The formal definition of that sigma notation is recursive:
$$\sum_{j=1}^1a_j=a_1,$$
$$\sum_{j=1}^{n+1}a_j=\sum_{j=1}^na_j+a_{n+1}.$$
Since the formal definition is recursive, anything you prove about it is going to involve induction at some point. Proving $\sum_{j=1}^Nc=Nc$ by induction is not hard.
