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$$\sum_{i=1}^n (i + 1) - \sum_{j=1}^n j $$

I cant really get my head around on how to simplify these sigma notations, any help would be appreciated.

Thanks

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    $\begingroup$ Just try it for modest $n$! Suppose $n=1$. Then try $n=2$. I expect a pattern will emerge. $\endgroup$
    – lulu
    Oct 23 '16 at 14:05
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    $\begingroup$ You could replace $j$ by $i$ and simplify. $\endgroup$ Oct 23 '16 at 14:08
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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 $$ \sum_{i=1}^n 1 = n $$

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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*}

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You have

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

so

$$\sum_{i=1}^n (i+1)-\sum_{j=1}^n j=\sum_{i=2}^{n+1} i-\sum_{j=1}^n j=n+1+\sum_{i=1}^n (i-i)+1=n+1-1=n.$$

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(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

$$\sum_{k=1}^n k=\sum_{1\le k\le n}k=\sum_{3\le h\le n+2}(h-2)=\sum_{h=3}^{n+2}(h-2)$$

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