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$$\sum_{x}^n x = (n + x) \frac{n - (1 - x)}{2}$$
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.

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  • $\begingroup$ sum of first n numbers? $\endgroup$
    – josf
    Commented Jan 22, 2017 at 9:41
  • $\begingroup$ No I mean x can be any number and just start adding from x to n $\endgroup$
    – user208685
    Commented Jan 22, 2017 at 9:43
  • $\begingroup$ Use $1+2+3+\cdots +n = \dfrac{n(n+1)}{2}$ $\endgroup$ Commented Jan 22, 2017 at 9:49
  • $\begingroup$ Sorry, wrong equation first time, edited $$\sum_{x}^n i = (n + x) \frac{n - (1 - x)}{2}$$ to $$\sum_{x}^n x = (n + x) \frac{n - (1 - x)}{2}$$ $\endgroup$
    – user208685
    Commented Jan 22, 2017 at 9:51

3 Answers 3

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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.

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  • $\begingroup$ Wait, that is what I got, so I'm right? $\endgroup$
    – user208685
    Commented Jan 22, 2017 at 11:43
  • $\begingroup$ No, the last term in your result is $n+x-1$ and not $n-x+1$ $\endgroup$
    – b00n heT
    Commented Jan 22, 2017 at 12:21
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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

$$\sum_{k=a}^b k=\sum\nolimits_a^{b+1}k\;\delta k=\frac{k(k-1)}2\bigg|_a^{b+1}=\frac{(b+1)b}2-\frac{a(a-1)}2$$

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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.

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  • $\begingroup$ I don't mean starting from 1, I mean starting from x $\endgroup$
    – user208685
    Commented Jan 22, 2017 at 9:53
  • $\begingroup$ But I use formula for starting from 1 because we know that formula, there is many ways you can prove this but you must use this what I wrote and @Steven Gregory in comment. $\endgroup$
    – josf
    Commented Jan 22, 2017 at 9:59

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