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 here. But some are not as easy as one might hope:
$$\sum_{i=0}^n\,i^k=\href{http://en.wikipedia.org/wiki/Stirling_numbers_of_the_first_kind}{\text{?}} \href{http://en.wikipedia.org/wiki/Faulhaber%27s_formula}{\text{(general formula for arbitrary $k$: hard)}}$$
$$\sum_{n=1}^\infty\,\frac1{n^s}=\href{http://en.wikipedia.org/wiki/Riemann_zeta_function}{\text{? (a very important but mysterious function)}}$$
