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

$\displaystyle\sum_{v=1}^nv=\dfrac{1}{2}\displaystyle\sum_{v=1}^n2v=\dfrac{1}{2}\displaystyle\sum_{v=1}^n\left((v+1)^2-v^2-1\right)=\dfrac{1}{2}\left((n+1)^2-(n+1)\right)-=\dfrac{1}{2}n(n+1)$
