Spivak's Calculus: Finding the formula for the sum of a series. In Spivak's Calculus, Chapter 2, the second problem of the set asks you to find a formula for the following series:
$$\sum_{i=1}^n(2i-1)$$ and $$\sum_{i=1}^n(2i-1)^2$$
Now, for the former, this was fairly straightforward: it sums to $i^2$ and I have a proof that I'm happy with. However, for the latter, I cannot identify a pattern in terms of $n$ to begin building a proof. The series proceeds $1^2 + 3^2 + 5^2 + 7^2 + \cdots + (2n-1)^2$, so the sum for the first few indices would be $1, 10, 35,$ and $84.$ Not only have I failed to come up with an expression of this in terms of $n$, but I'm not even sure what I should be considering to lead me to such a formula.

$\sum_\limits{k=1}^n (2k - 1)^2 = \sum_\limits{k=1}^n 4k^2 - \sum_\limits{k=1}^n 4k + \sum_\limits{k=1}^n 1 = 4\left(\sum_\limits{k=1}^n k^2\right) -  2n(n+1) + n$
You need a way to find $\sum_\limits{k=1}^n k^2$
There are lots of other ways to derive this... but this one is nice.
Write the numbers in a triangle
$\begin {array}{} &&&1\\&&2&&2\\&3&&3&&3\\&&&\vdots&&\ddots\\n&&n&\cdots\end{array}$
The sum of each row is $k^2.$ And, the sum of the whole trianglular array is $\sum_\limits{k=1}^n k^2$
Take this triangle and rotate it 60 degrees clockwise and 60 degrees counter clockwise and sum the 3 triangles together.
$\begin {array}{} &&&1\\&&2&&2\\&3&&3&&3\end{array}+\begin {array}{} &&&n\\&&n-1&&n\\&n-2&&n-1&&n\\\end{array}+\begin {array}{} &&&n\\&&n&&n-1\\&n&&n-1&&n-2\end{array}$
And we get:
$\begin {array}{} &&&2n+1\\&&2n+1&&2n+1\\&2n+1&&2n+1&&2n+1\\&&&\vdots&&\ddots\end{array}$
What remains how many elements are in the triangular array?  $\sum_\limits{k=1}^n k$
$3\sum_\limits{k=1}^n k^2 = (2n+1)\sum_\limits{k=1}^n k\\
\sum_\limits{k=1}^n k^2 = \frac {n(n+1)(2n+1)}{6}$
Another way to do it is: 
$\sum_\limits{k=1}^n (k+1)^3 - k^3 = (n+1)^3-1\\
\sum_\limits{k=1}^n (3k^2 + 3k + 1)\\
\sum_\limits{k=1}^n 3k^2 + \sum_\limits{k=1}^n 3k + \sum_\limits{k=1}^n 1\\
\sum_\limits{k=1}^n 3k^2 + 3\frac {n(n+1)}{2} + n = n^3 + 3n^2 + 3n \\
3\sum_\limits{k=1}^n k^2 = n^3 + \frac 32 n^2 + \frac 12 n\\
\sum_\limits{k=1}^n k^2 = \frac 16 (n)(n+1)(2n+1)$
