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

Question

How to come from this equation to

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

this equation:

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

Thanks in advance for your answer!

Have you tried getting a common denominator in the first equation and seeing if you can get it to factor into the second equation?
That is, the $(n+1)$ in the second summand of the first displayed equation is equal to $\frac{2(n+1)}{2}$.
The sum of the first n odd numbers is n squared, use some algebraic manipulation to get your sum, and its n(n+1)/2
You can't, as the problem is stated. The second equality should be $\sum_{i=0}^{n+1}\;k$.
As stated this is not correct. $\sum_{i=0}^nk$ should be replaced by $\sum_{k=0}^{n+1}k$ everywhere.

Shaktal · Accepted Answer · 2012-12-10 21:18:37Z

up vote 2 down vote accepted

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

edited Dec 10 '12 at 21:18

Shaktal
3,2831420

answered Dec 10 '12 at 20:36

omid saba
816

1 Answer