Finding the closed form of a sum

Question

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.

Answer 1

$$\sum_{n=4}^x(x-n)$$ 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 $$\frac N2(a+l)$$

Answer 2

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

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

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