# Finding the closed form of a sum

I would like to find the closed form of the sum $\sum_{n = 4}^{x}(x - n)$. I believe that the derivative is $x - 4$, but when I take the integral of that and graph it, the sum and $\frac{x^2}{2} + 4x$ are certainly not the same. Any help would be appreciated, as I have no idea how to proceed.

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$$\sum_{n=4}^x(x-n)$$ is an Arithmetic Series with the common difference $=1$

as the $r(0\le r\le -n-x+4)$th term$(T_r)$ is $x-n-r-4$

So, $T_{r+1}-T_r=1$

the first term being $x-4$ and the last being $x-x=0$ and the number of terms is $\displaystyle (x-4)-(x-x)+1=x-4+1$

Now, the sum of $N$ term with the first & the last term being $a,l$ is $$\frac N2(a+l)$$

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So the closed form would be \$)\frac{x/2})(x-4)? – recursive recursion Feb 14 '14 at 18:20
@recursiverecursion, it should be $$\frac{x-3}2(x-4+0)$$ – lab bhattacharjee Feb 14 '14 at 18:22
Thank you! great answer, didn't read it correctly the first time. – recursive recursion Feb 14 '14 at 18:24
@recursiverecursion, my pleasure. Hope I could make the idea clear – lab bhattacharjee Feb 14 '14 at 18:25
This was the last part of deriving the formula for the number of diagonals in a polygon. I looked just now, and my formula's right :) – recursive recursion Feb 14 '14 at 18:34

$$\sum_{n=4}^x{(x-n)}=\sum_{n=4}^x{x}-\sum_{n=4}^x{n}$$ $$=(x-3)x+(4+5+6+...+x)$$

$$=(x^2-3x)-\left(\frac{x(x+1)}{2}-6\right)$$ $$=x^2-3x-\frac{x^2}{2}-\frac{x}{2}+6$$ $$=\frac{x^2}{2}-\frac{7x}{2}+6$$

$$=\frac{1}{2}(x^2-7x+12)$$ $$=\frac{1}{2}(x-4)(x-3)$$

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