Induction Pyramid Question Volume I am starting to learn induction proofs and I ran into issues trying to prove this one question after working on it for a couple of hours tonight. I understand why it's true (relating pyramid height & squares left), but when trying to prove it I have been hearing stuff like "getting rid of the bottom row" but that just makes no sense to me. 
$$\sum_{k=1}^n k^2 = \sum_{k=1}^n (n+1-k)(2k-1)$$
(My TA recommended using math exchange as a way to get answers for these questions, I am sorry unfortunately I cannot yet insert images directly) 
 A: The comment "getting rid of the bottom row" probably refers to some geometric way to think about the problem, which is not going to make much sense unless someone explicitly draws out the geomteric picture for you. For now, if you want a purely algebraic way to prove this through induction, we can use the following.
First, we set the proposition $P(n)$ to be the claim that
$$\sum_{i=1}^n i^2=\sum_{i=1}^n(n+1-i)(2i-1).$$
To prove a claim via induction, there are two steps. First, we show the base case, i.e. that $P(1)$ is true. Then, we assume for some $k$ that $P(k)$ is true, and under this assumption, we show that $P(k+1)$ is true, which proves the claim $P(n)$ for all $n$.
The base case here is just the claim that $1^2=(1+1-1)(2\cdot 1-1)$, which is obviously true, since there is only one term in both sums. Now we suppose that for some $k$, we indeed have
$$\sum_{i=1}^k i^2=\sum_{i=1}^k(k+1-i)(2i-1).$$
We want to show that
$$\sum_{i=1}^{k+1} i^2=\sum_{i=1}^{k+1}(k+2-i)(2i-1).$$
Of course, the whole point of induction is to make use of the inductive hypothesis. So we look at the right hand side and manipulate it so that we can make use of the hypothesis. Write
$$\begin{split}\sum_{i=1}^{k+1}(k+2-i)(2i-1)&=\left[\sum_{i=1}^k(k+2-i)(2i-1)\right]+(2k+1)\\
&= \left[\sum_{i=1}^k\left((k+1-i)+1\right)(2i-1)\right]+(2k+1)\\
&= \left[\sum_{i=1}^k(k+1-i)(2i-1)+\sum_{i=1}^k1\cdot(2i-1)\right]+(2k+1).
\end{split}$$
Now, notice that the first term in the sum is exactly $\sum_{i=1}^k i^2$, by the induction hypothesis. The second term can be evaluated nicely since it is the sum of an arithmetic progression. Can you finish the proof now, by showing that the second and third term sum to $(k+1)^2$ and hence that this sum is equal to $\sum_{i=1}^{k+1}i^2$?
