How would I simplify this summation: $\sum_{i=1}^n (i + 1) - \sum_{j=1}^n j $ $$\sum_{i=1}^n (i + 1) - \sum_{j=1}^n j $$  
I cant really get my head around on how to simplify these sigma notations, any help would be appreciated.
Thanks
 A: You can “reorganize” the first summation:
$$
\sum_{i=1}^n(i+1)=\sum_{i=1}^n i+\sum_{i=1}^n1
$$
Since
$$
\sum_{i=1}^n i = \sum_{j=1}^n j
$$
you remain with
$$
\sum_{i=1}^n 1 = n
$$
A: Another variation which might be helpful.

\begin{align*}
\sum_{i=1}^n (i + 1) - \sum_{j=1}^n j&=\sum_{i=1}^n (i + 1) - \sum_{i=1}^n i\\
&=\sum_{i=1}^n (i + 1-i)\\
&=\sum_{i=1}^n 1\\
&=n
\end{align*}

A: You have 
$$\sum_{i=1}^n (i+1)=\sum_{i=2}^{n+1} i$$
so
$$\sum_{i=1}^n (i+1)-\sum_{j=1}^n j=\sum_{i=2}^{n+1} i-\sum_{j=1}^n j=n+1+\sum_{i=1}^n (i-i)+1=n+1-1=n.$$
A: (Too long for a comment)
As @lulu said you can investigate the behavior of the sum by hand but you can use a bit of algebra before to simplify something. By example observe that
 $$\sum_{k=1}^n (k+1)=\left(\sum_{k=1}^n k\right)+\left(\sum_{k=1}^n 1\right)=\left(\sum_{k=1}^n k\right) + n$$ and $$\sum_{k=1}^n k=\sum_{j=1}^n j=1+2+3+\cdots+n$$
Observe too that a summation can be written as
$$\sum_{k=1}^n k=\sum_{1\le k\le n}k$$
Then you can manipulate easily the inequality $1\le k\le n$ if you need to change the variable $k$ by, for example, $h=k+2$
$$1\le k\le n\iff 1+2\le k+2\le n+2\iff 3\le h\le n+2$$
Then you can rewrite
$$\sum_{k=1}^n k=\sum_{1\le k\le n}k=\sum_{3\le h\le n+2}(h-2)=\sum_{h=3}^{n+2}(h-2)$$
