is my simplification of sigma correct? $$\sum_{x}^n x = (n + x) \frac{n - (1 - x)}{2}$$
so far my calculation has proved this formula will work.
something tells me that the chance that my calculation
and logic is legit is pretty slim. Can you disapprove this equation?
I'm also presuming that x is a whole number.
 A: The correct formula should read
$$\sum_{k=x}^nk.$$
That said, using
$$\sum_{k=1}^mk=1+\cdots+ m=\frac{m\cdot(m+1)}2$$
and rewriting your sum as
$$\sum_{k=x}^nk=\sum_{k=1}^nk-\sum_{k=1}^{x-1}k.$$
one obtains
\begin{align}\sum_{k=x}^nk& = \color{red}{\sum_{k=1}^nk}-\color{blue}{\sum_{k=1}^{x-1}k}\\
& = \color{red}{\frac{n\cdot(n+1)}2}-\color{blue}{\frac{(x-1)\cdot((x-1)+1)}2}\\ & =\frac{n\cdot(n+1)}2-\frac{x\cdot(x-1)}2\\ & = \frac{n^2-x^2+n+x}{2}\\
\\ & = \frac{(n+x)(n-x)+(n+x)\cdot 1}{2}
\\ &=\frac{(n+x)(n-x+1)}{2}
\end{align}
which is almost what you got.
A: Supposing that you are asking for the closed form of
$$\sum_{k=a}^b k$$
then from finite calculus we have that
$$\sum k\;\delta k=\frac{k^{\underline 2}}{2}+C=\frac{k(k-1)}2+C\tag{1}$$
where the expression $k^{\underline m}:=\prod_{j=0}^{m-1} k-j$ is a falling factorial, and $C$ is a function such that $C(n)=\ell$ for some constant $\ell$ when $n\in\Bbb N$.
Then taking limits in (1) we have
$$\sum_{k=a}^b k=\sum\nolimits_a^{b+1}k\;\delta k=\frac{k(k-1)}2\bigg|_a^{b+1}=\frac{(b+1)b}2-\frac{a(a-1)}2$$
A: Ok than the sum is equal to the (sum of $1$ to $n$) - (sum of $1$ to $x$).
And this two sums we can easily calculate by using Gauss idea - https://en.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_%E2%8B%AF

Example: $1+2+3+4+5+6+7$ see that 1+7 = 2+6 = 3+5 ... using this idea you can prove formula from the given link.
