When my teacher gives me a question involving summation notation, do they expect us to calculate it by hand? Assuming we don't have a calculator that can do summation notation. My class is not up to summation yet, but I'm asking a question involving this concept because I'm not all that experienced using it. My calculator can do summation but that's because I programmed it to. But when we actually catch up to this concept, are we to calculate it by hand?
Take, for example:
$\displaystyle \sum_{i = 0}^3 2i + 1$
Do you think they will tell us for the $4$ values of $i$ (from $0$ to $3$), do the following operation, or is there some other faster way of doing it?
Sorry if this is a silly question, this has just been on my mind...
 A: Ask your teacher. Mileage may vary. Just make sure you know you have the right answer, whichever method you use. In this case, there is actually a very nice formula: $$\sum_{k=0}^{n-1}{2k+1}=n^2$$ and in your case, $n=4$. You can also think of this as $4=3-0+1$ (the number of terms being added) times the average value of each term, which is $2x+1$ for $x=\frac{0+3}{2}$. And the formula works because $(n+1)^2=n^2+(2n+1)$. I have formulated the summation going from $0$ to $n-1$ because this has $n$ terms. Perhaps you would be more comfortable with the $(n+1)$-term version: $$\sum_{k=0}^{n}{2k+1}=(n+1)^2$$ Or perhaps you could generalize this to $$\sum_{k=a}^{b}{2k+1}=(b+1)^2-a^2=(b+a+1)(b-a+1)$$ for $a\le b$. It takes a bit of getting used to, and even then a bit of care working with summations, but it is time very well spent, in my opinion. In fact, there are some really interesting generalizations:
$$\sum_{i=1}^{n}\,i=\frac{n(n+1)}{2}$$
$$\sum_{i=1}^{n}\,\frac{i(i+1)}{2}=\frac{n(n+1)(n+2)}{1\cdot2\cdot3}$$
and some other ones I'm sure you'll love:
$$\sum_{i=0}^{n-1}\,a^i=\frac{a^n-1}{a-1}\qquad\text{for }a\ne1$$
$$\sum_{n=1}^\infty\,\frac1{n^2}=\frac{\pi^2}{6}$$
More in the "Capital-sigma notation" section of the Wikipedia article with the same name. But some are not as easy as one might hope:


*

*$$\sum_{i=0}^n\,i^k=$$

*general formula for arbitrary $k$: (hard)

*$$\sum_{n=1}^\infty\,\frac1{n^s}=$$
(a very important but mysterious function)
