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I've seen a couple of different variation of the summation signs, and can anyone please clarify them to me since it's pretty confusing. The one that I can understand is the most common onse that we've all learned in calculus class, for example: $$\sum^\infty_{i=1}\frac{1}{i}=\infty$$ The rest, is very confusing, all the numbers is on the buttom with nothing on the top, for example, the multinomial theorem:$$(x_1+x_2+...+x_k)^n=\sum_{a_1,a_2,...,a_k}{n\choose{a_1,a_2,...,a_k}}x_1^{a_1}x_2^{a_2}...x_k^{a_k}$$ or, another formula to calculate $(x+y)^n$ when $n$ is not a natural number:$$(1+x)^n=\sum_{i\geq0}{n\choose{i}}x^i$$ or something else like: $$\sum_{i=0}_{i\ odd}^n...$$$$\sum_{a_1=n}^n...$$

Can someone please explain what does all these summation signs mean? Ty.

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TZakrevskiy’s answer covers all of your examples except the one from the multinomial theorem. That one is abbreviated, and you simply have to know what’s intended or infer it from the context: the summation is taken over all $k$-tuples $\langle a_1,\ldots,a_k\rangle$ of non-negative integers satisfying the condition that $a_1+a_2+\ldots+a_k=n$. If the condition were written out in full, the summation would look like

$$\huge\sum_{{a_1,\ldots,a_k\in\Bbb N}\atop{a_1+\ldots+a_k=n}}\ldots\;.$$

(Note that my $\Bbb N$ includes $0$.)

Added example: Let $n=2$ and $k=3$. The ordered triples $\langle a_1,a_2,a_3\rangle$ that satisfy $a_1+a_2+a_3=2$ are:

$$\begin{array}{ccc} a_1&a_2&a_3\\ \hline 0&0&2\\ 0&2&0\\ 2&0&0\\ 0&1&1\\ 1&0&1\\ 1&1&0 \end{array}$$

Thus, the sum in question is

$$\begin{align*} \binom2{0,0,2}x_1^0x_2^0x_3^2&+\binom2{0,2,0}x_1^0x_2^2x_3^0+\binom2{2,0,0}x_1^2x_2^0x_3^0\\ &+\binom2{0,1,1}x_1^0x_2^1x_3^1+\binom2{1,0,1}x_1^1x_2^0x_3^1+\binom2{1,1,0}x_1^1x_2^1x_3^0\;. \end{align*}$$

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  • $\begingroup$ Brian, what does this sum actually represents, I'm still confused, what does the addition sign stands for? Can you explain further what does a sum that's like $\sum_{a_1+a_2+...=n}$ means? $\endgroup$ – Commander Shepard Sep 25 '13 at 2:11
  • $\begingroup$ @CommanderShepard: I’ve added a small example; see if it helps. The condition $a_1+\ldots+a_k=n$ is a condition that we demand the indices satisfy: we have one term for every combination of indices $a_1,\ldots,a_k$ satisfying the conditions under the summation sign. $\endgroup$ – Brian M. Scott Sep 25 '13 at 2:21
  • $\begingroup$ Thanks for the detailed explanation! Now I totally understand it, thanks professor! $\endgroup$ – Commander Shepard Sep 25 '13 at 2:26
  • $\begingroup$ @CommanderShepard: You’re welcome! $\endgroup$ – Brian M. Scott Sep 25 '13 at 2:26
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When we take the sum of elements, we don't want to write an infinite expression, like in the case of series $$\sum_{i=1}^\infty\frac{1}{i^2}=\frac{\pi^2}{6};$$ we don't want to write sums with the number of terms being a variable, like in the case $$\sum_{i=0}^n\binom{n}{i}=2^n.$$ And we don't want to write anything really long or nasty-looking, like $$\sum_{i=0}^{99} \frac{1}{2^i}=2(1-2^{-100}).$$

We want to give some sort of formula for the the terms of our sum - i.e. make them a function of some index - and then describe the possible values of that index. Then the notation conventions come into play. Unless otherwise specified, the index variable is a natural number (I take $0$ as a natural number); so, the sums

$$\sum_{i=0}^{\infty}a_i,\quad \sum_{i\ge 0}a_i,\quad \sum_{i \in \Bbb N}a_i, \quad \sum_{\Bbb N} a_i$$ represent exactly the same thing.

If we use the notation $\sum_{i=j}^{k}a_i$, it means that $i$ runs from $j$ to $k$ with the step of $1$; we can also ask something specific from our index, as in your example $$\sum_{i=0}_{i\ odd}^n a_i,$$where the index $i$ takes only odd values in the interval $[0,n]$. To illustrate, we write $$\sum_{i=1}_{i\ odd}^7\frac 1i = 1 +\frac 13 +\frac 15+\frac 17.$$

I hope this wall of text helps to clear your confusion.

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  • $\begingroup$ How about the ones in multinomial theorem, what do represent? Like the sum that's $\sum_{a_1,a_2,...}$? $\endgroup$ – Commander Shepard Sep 25 '13 at 2:01
  • $\begingroup$ @CommanderShepard sorry, forgot about it. Anyway, Brian has already covered it. $\endgroup$ – TZakrevskiy Sep 25 '13 at 2:14

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