$\sum_{i=1}^n i$ is the same as $\frac{n(n+1)}{2}$. Can someone explain how the sigma notation is converted to this? I'm trying to figure out if there's a way to convert $\sum_{i=1}^n i+(x-1)$.

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    $\begingroup$ Be careful with the notation. $\sum_{i=1}^ ni+(x-1)$ is just $\frac{n(n+1)}{2}+(x-1)$, whereas $\sum_{i=1}^ n(i+x-1)$ is $\frac{n(n+1)}{2}+n\cdot (x-1)$. $\endgroup$ – Hagen von Eitzen Sep 2 '17 at 19:01
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    $\begingroup$ The fact expressed by saying $\displaystyle \sum_{i=1}^n i = \frac{n(n+1)} 2$ is also expressed by saying $\displaystyle 1 + 2 + 3 + \cdots + n = \frac{n(n+1)} 2,$ without the sigma notation. Is your question how $1+2+3+\cdots + n$ is converted to $\displaystyle \frac {n(n+1)} 2,$ or is it something about dealing with the sigma notation? $\qquad$ $\endgroup$ – Michael Hardy Sep 2 '17 at 23:34

You convert notations by using the definition of the sigma notation. Let us say we have a (finite, for simplicity) set $\{a_1, a_2, \ldots , a_n\}$. If we want to sum all of its elements, then we can write $\sum_{i=1}^n a_i$, which simply tells us to first insert $i=1$ into what is inside the sigma notation, then add what you get when you insert $i=2$, then add what you get when you insert $i=3$, et cetera, all the way up to $i=n$. In symbols, $$\sum_{i=1}^n a_i = a_1+a_2+\cdots+a_n.$$ You can also do this with an index set $I=\{1,2, \ldots, n\}$; in that case, we write $$\sum_{i \in I} a_i = a_1+a_2+\cdots+a_n.$$

Your first example $\sum_{i=1}^n i$ should be converted as $1+2+3+\cdots+n$, because you just insert $i=1,2,3,\ldots,n$ into the formula inside the sigma notation. It is true that this equals $n(n+1)/2$, but one must prove that! The formal way of doing this is with the principle of mathematical induction, but you can see it is true without any machinery at all (c.f. this famous story of how Gauss supposedly summed the numbers from one to one-hundred).

In the case of $\sum_{i=1}^n i+(x-1)$, the direct translation of notations is as follows: $$\sum_{i=1}^n i+(x-1)=[1+(x-1)]+[2+(x-1)] + \cdots + [n+(x-1)],$$ where you can of course remove the parentheses and brackets as you wish.

Or, maybe you mean to write $\sum_{i=1}^n(i) +(x-1)$ with the $x-1$ term outside the sigma notation. In that case, we have

$$\sum_{i=1}^n(i) +(x-1)=(1+2+\cdots+n)+(x-1)=\frac{n(n+1)}{2}+(x-1),$$ where the final equality is the result of the aforementioned theorem on the sum of the first $n$ natural numbers.


This is the proof that I am most familiar with.

Let $S$ be equal to $\sum_{i=1}^n i = 1 + 2 + 3 \cdots + (n-1) + n$

Then $S+S = \Bigl( 1 + 2 + 3 \cdots + (n-1) + n\Bigr) + \Bigl(n + (n-1) + \cdots + 3 + 2 + 1\Bigr)$

See that we get $2S = \underbrace{(n+1) + (n+1) + \cdots + (n+1)}_{n \text{ times}} = n(n+1)$

Or $2S = n(n+1)$

Division by $2$ gives us $S = \frac{n(n+1)}{2}$

As for your second question, I must agree with what Hagen von Eitzen said in the comments, as depending on how you write the formula, it could give slightly varying results.

Note however that it will be easy to produce a formula as summation breaks apart across addition.

$\sum_{i=1}^n \Bigl(i+(x-1)\Bigr) = \sum_{i=1}^n i + \sum_{i=1}^n (x-1)$


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