What does $\prod\limits_{k=-2}^{11}(15-3k)$ mean--and how might I compute it? $$\prod_{k=-2}^{11}(15-3k)=\;?$$
I'm new to this and have not seen this notation before. Can anyone explain to me what this is called and how to solve or compute it?
 A: Note on notation: As others have mentioned, the $\prod$ symbol stands for "product" and is actually typeset by $\prod$; similarly, the $\sum$ symbol stands for "sum" and is actually typeset by $\sum$. You may have seen the $\Sigma$-notation before, but let's consider the $\Pi$-notation because that's what you are really interested in here. The Wikipedia page on $\Pi$-notation is good and should get you started. They give the example
$$
\prod_{k=1}^4 k = 1\cdot 2\cdot 3\cdot 4 = 24.\tag{1}
$$
Can you see how that worked? You multiply everything together for the values $k=1,2,3,4$. Similarly, for $\prod_{k=1}^n k$, you would multiply everything together for the values $k=1,2,\ldots,n$. So let's consider a more interesting example:
$$
\prod_{k=1}^4 (k+2) = \underbrace{(\color{red}{1}+2)}_{\color{red}{k\,=\,1}}\cdot\underbrace{(\color{red}{2}+2)}_{\color{red}{k\,=\,2}}\cdot\underbrace{(\color{red}{3}+2)}_{\color{red}{k\,=\,3}}\cdot\underbrace{(\color{red}{4}+2)}_{\color{red}{k\,=\,4}} = 3\cdot 4\cdot 5\cdot 6 = 360.\tag{2}
$$
Make sense? Now let's consider your problem.
Your problem:
We are tasked with computing
$$
\prod_{k=-2}^{11} (15-3k).
$$
Based on what you saw in $(1)$ and $(2)$, do you see how you might compute this? You can write out all of the terms, but this proves to be a fruitless endeavor as other answerers have pointed out. Here's why:
$$
\prod_{k=-2}^{11} (15-3k) = \underbrace{[15-(-6)]}_{k\,=\,-2}\cdot\underbrace{[15-(-3)]}_{k\,=\,-1}\cdot\ldots\cdot\color{red}{\underbrace{\overbrace{(15-15)}^{\color{blue}{=\;0}}}_{k\,=\,5}}\cdot\ldots\cdot\underbrace{(15-33)}_{k\,=\,11}.\tag{3}
$$
Simplifying $(3)$ a bit, you can see that
$$
\prod_{k=-2}^{11} (15-3k) = 21\cdot 18\cdot\ldots\cdot\color{red}{0}\cdot\ldots\cdot -18=0
$$
because any number multiplied by zero is zero. This is what Jasper Loy's answer immediately noted. Do you see it now?            
A: When $k=5$, $15-3k=0$, so the product is $0$.
A: The symbol $\prod$ indicates a product, in a similar fashion as the $\sum$ for sums. In this case, the formula works out as $$(15 - 3(-2)) \cdot (15 - 3(-1)) \cdot (15 - (3\cdot0)) ...$$
A: just as we write $$a_1 + a_2 + a_3 + \cdots + a_{10} = \sum_{j=1}^{10}a_j,$$  the product is written as $$a_1a_2\cdots a_{10} = \prod_{j=1}^{10} a_j $$ 
so $$\prod_{k = -2}^{11}(15-3k) = 21 \times 18 \times\cdots\times 0 \cdots(-18) = 0 $$
A: $$\prod_{k=-2}^{11}(15-3k)=21\cdot18\cdot15\cdot12\cdot9\cdot6\cdot3\cdot0\cdot....\cdot(-18)=0$$
