I think the idea is to try to find the formulas yourself. For instance, the first few examples would be
$\displaystyle \sum_{k = 1}^{2} (2k - 1) = (2(1) -1) + (2(2) - 1) = 1 + 3 = 4$
$\displaystyle \sum_{k = 2}^{4} (2k - 1) = (2(2) -1) + (2(3) - 1) + (2(4) - 1) = 3 + 5 + 7 = 15$
$\displaystyle \sum_{k = 3}^{6} (2k - 1) = (2(3) -1) + (2(4) - 1) + (2(5) - 1) + (2(6) - 1)$ $\displaystyle = 5 + 7 + 9 + 11 = 32$
You may not be able to see the series from this but I'll give you a hint: $\displaystyle S(n) = an^2 + bn + c$. You've got three equations and three unknowns.
-Dan
It always help to look at different small values for n.
$\displaystyle \sum_{k=1}^{2*1}(2k - 1) = \sum_{k=1}^2(2k - 1) =$
$ \{(2 * 1) - 1\} + \{(2 * 2) - 1\} = 1 + 3 = 4 = 2^2 = (2 * 1)^2 - 0.$
$\displaystyle \sum_{k=2}^{2*2}2k - 1) = \sum_{k=2}^4(2k - 1) =$
$\{(2 * 2) - 1\} + \{(2 * 3) - 1\} + \{(2 * 4) - 1\} = 3 + 5 +7 = 15 = 16 - 1 = (2 * 2)^2 - 1.$
$\displaystyle \sum_{k=3}^{2*3}(2k - 1) = \sum_{k=3}^6(2k - 1) =$
$\{(2 * 3) - 1\} + \{(2 * 4) - 1\} + \{(2 * 5) - 1\} +\{(2 * 6) - 1\} = 5 + 7 + 9 + 11 = 32 = (2 * 3)^2 - 4.$
Does this remind you of anything?
I'll give you another hint.
$\displaystyle \sum_{i=1}^m(2i - 1) = what?$
Not clever enough to do what? He and I just did some arithmetic first. Nothing clever about that.
Then topsquark, who is cleverer than I, saw that a quadratic would do the trick without any apparent intermediate steps. But I saw that this sum is related to the sum of the first p odd numbers and so started looking for squares too.
He gave you an analytic approach. I gave you a pattern recognition approach. What is it that you are not seeing?
$\displaystyle \sum_{k = n}^{2n} (2 k - 1) = \text{?}$.
Define $\displaystyle f(m) = \sum_{k = 1}^{m} (2 k - 1)$.
Then $\displaystyle \sum_{k = n}^{2n} (2 k - 1) = f(2n) - f(n-1)$, because always have (for $\displaystyle n \ge 2$) that..... $\displaystyle \sum_{k = 1}^{n-1} a_k + \sum_{k = n}^{2n} a_k = \sum_{k = 1}^{2n} a_k$.
So to find $\displaystyle \sum_{k = n}^{2n} (2 k - 1)$, it suffices to find $\displaystyle f(m)$.
But $\displaystyle f(m) = \sum_{k = 1}^{m} (2 k - 1) = \sum_{k = 1}^{m} (2 k) + \sum_{k = 1}^{m} (- 1) = 2 \sum_{k = 1}^{m} k - \sum_{k = 1}^{m} (1)$.
Thus $\displaystyle f(m) = 2 \left( \sum_{k = 1}^{m} k \right) - m$.
I bet you've seen $\displaystyle \sum_{k = 1}^{m} k$ before, and if so, you can find $\displaystyle f(m)$, and so you can get a formula for $\displaystyle \sum_{k = n}^{2n} (2 k - 1) = f(2n) - f(n-1)$.
I would work from the sum of the first $n$ natural numbers, $\sum_{k=1}^n k$, which, as johnsomeone says, you ought to have seen before. Given that, you have $$\sum_{k=1}^n (2k - 1) =\sum_{k=1}^{2n} k - \sum_{k=1}^n 2k$$ which just says that the sum of the first $n$ odd numbers is the sum of the first $2n$ natural numbers less the sum of the first $n$ even numbers.
Having said that, if you work out the first few terms $\sum_{k=1}^n (2k - 1)$, you should very definitely see a pattern.
You should know that the sum of the first N odd integers is N^2, so
$\displaystyle \begin{align*} \sum_{k = 1}^{N}{ \left( 2k - 1 \right) } &= N^2 \\ \\ \sum_{k = 1}^{2n} \left( 2k - 1 \right) &= \left( 2n \right) ^2 \\ &= 4n^2 \\ \\ \sum_{k = 1}^{n - 1} \left( 2k - 1 \right) &= \left( n - 1 \right) ^2 \\ &= n^2 - 2n + 1 \\ \\ \sum_{ k = n}^{2n} \left( 2k - 1 \right) &= \sum_{ k = 1 }^{2n} \left( 2k - 1 \right) - \sum_{k= 1}^{n - 1} \left( 2k - 1 \right) \\ &= 4n^2- \left( n^2 - 2n + 1 \right) \\ &= 3n^2 + 2n - 1 \end{align*}$
Now you could try induction if you want...