valid proof of series $\sum \limits_{v=1}^n v$ $$\sum \limits_{v=1}^n v=\frac{n^2+n}{2}$$
please don't downvote if this proof is stupid, it is my first proof, and i am only in grade 5, so i haven't a teacher for any of this 'big sums'
proof:
if we look at $\sum \limits_{v=1}^3 v=1+2+3,\sum \limits_{v=1}^4 v=1+2+3+4,\sum \limits_{v=1}^5 v=1+2+3+4+5$
i learnt rainbow numbers in class three years ago, so i use that knowlege here:
$n=3,1+3=4$ and $2$.
$n=4,1+4$ and $2+3$
$n=5,1+5$ and $2+4$ and $3$
and more that i have done on paper that i don't wanna type.
we can see from this for the odd case that we have $(n+1)$ added together moving in from the outside, so we get to add $(n+1)$ to the total $\frac{(n-1)}2$ times plus the center number, which is $\frac{n+1}2$.. giving $\frac{n-1}2(n+1)+\frac{n+1}2=\frac{(n+1)(n-1)}{2}+\frac{n+1}{2}$ and i can get $\frac{n^2-1}2+\frac{n+1}2=\frac{n^2+n}2$ which is what we want.
so odd are proven.
for even we have a simplier problem: we have $n+1$ on each pair of numbers going in. since we are even numbers, we have $1+n=n+1$ , with $n$ even, $2+(n-1)=n+1$ and we can see this is good for all numbers since we increase one side by one and lower the other by 1. so we get $\frac{n}2$ times $n+1$ gives $\frac{n^2+n}{2}$
thus is proven for all cases. thus is is proven
 A: I'm not familiar with "rainbow numbers", and I'm afraid I can't follow every step of your proof. But if you're just looking for a very elementary proof of this, here's the easiest one I can think of:
Write the sum forwards:
$S_n = 1 + 2 + 3 + ... + n$
and then backwards:
$S_n = n + (n-1) + (n-2) + ... + 1$
and then sum term by term to get:
$2S_n = (n+1) + (n+1) +... + (n+1)$
where there are exactly $n$ of those terms.
So $2S_n = n(n+1)$
and $S_n = \frac{1}{2}n(n+1)$.
My apologies if this doesn't answer your question. I just thought you might want a nice elementary method to approach this (and it looks like less work than splitting into cases, etc.)
A: $\displaystyle\sum_{v=1}^nv=\dfrac{1}{2}\displaystyle\sum_{v=1}^n2v=\dfrac{1}{2}\displaystyle\sum_{v=1}^n\left((v+1)^2-v^2-1\right)=\dfrac{1}{2}\left((n+1)^2-(n+1)\right)-=\dfrac{1}{2}n(n+1)$
A: While most of the proofs that you will see are algebraic, sometimes it is useful to get a geometric view of the problem. I've always preferred getting multiple perspectives to give me deeper understanding of the problem at hand.
In the image, there are 5 different views of the problem. The first one has $(n+1)^2 - (n+1)$ cookies arranged in a square, with the diagonal removed. The second one arranges $n^2$ pizza into a square and then cuts the square in half. The third view arranges two sets of cookies into triangles to form a single rectangle. The fourth view we arrange squares into $n$ Ls that fit together to form a rectangle. Lastly, we have $n+1$ computers on a network that connects every computer directly to every other computer.  
As an exercise, try cutting the middle row of pizzas in half horizontally, and rearrange the triangle of pizzas into a rectangle.

