<s>[INST] Why is $\sum_{m=0}^k \frac{k^2+k}{2} = \frac{(k^2+k)(k+1)}{2}$? I recieved an answer like this :
$S= \displaystyle \sum_{m=0}^k \dfrac{k^2+k}{2} - k\displaystyle \sum_{m=0}^k m + \displaystyle \sum_{m=0}^k m^2=\dfrac{(k^2+k)(k+1)}{2}-\dfrac{k\cdot k(k+1)}{2}+\dfrac{k(k+1)(2k+1)}{6}=k(k+1)\left(\dfrac{k+1}{2} - \dfrac{k}{2}+\dfrac{2k+1}{6}\right)= \dfrac{k(k+1)(k+2)}{3}$
But I don't understand this equation $\sum_{m=0}^k \dfrac{k^2+k}{2} = \dfrac{(k^2+k)(k+1)}{2}.$ 
Could anyone explain me please?
 [/INST] Since $\displaystyle \sum_{x=0}^n C =(n+1) C$. you are summing constant $n+1$ times.
Note that you are summing over $m$ and $\dfrac {k^2+k}{2}$ is independent of $m$.
Note that $\displaystyle \sum_{x=0}^n C = \underbrace{C+C+ \dots +C}_{k+1 \text{times}} = (n+1) C$
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