# How to come to $\sum_{i=0}^n k = {(n+1)\cdot (n+2)\over 2}$ this equation?

How to come from this equation to

$$\sum_{i=0}^n k = {n(n+1)\over 2} + (n+1)$$

this equation:

$$\sum_{i=0}^n k = {(n+1)(n+2)\over 2}$$

That is, the $(n+1)$ in the second summand of the first displayed equation is equal to $\frac{2(n+1)}{2}$. –  user1551 Dec 10 '12 at 20:35
You can't, as the problem is stated. The second equality should be $\sum_{i=0}^{n+1}\;k$. –  Rick Decker Dec 10 '12 at 21:24
As stated this is not correct. $\sum_{i=0}^nk$ should be replaced by $\sum_{k=0}^{n+1}k$ everywhere. –  Dave Dec 10 '12 at 21:26
$$\sum_{i=0}^{n}k=\frac{n(n+1)}{2}+(n+1)=\frac{n(n+1)}{2}+\frac{2(n+1)}{2}=\frac{n(n+1)+2(n+1)}{2}=\frac{(n+2)(n+1)}{2}$$