A cruel and fickle king

The number of cubic stones in the pyramid will be the sum of the first 100 squares of odd numbers. We can write and simplify the finite series:

$\sum\limits_{k=1}^{100}{(2k-1)^2}=\sum\limits_{k=1}^{100}{(4k^2-4k+1)}=4\sum\limits_{k=1}^{100}{k^2}-4\sum\limits_{k=1}^{100}{k}+100$

These series can be calculated using the sum of squares formula and the sum of natural numbers formula:

$\sum\limits_{k=1}^{n}{k^2}=\frac{n(n+1)(2n+1)}{6}$

$\sum\limits_{k=1}^{n}{k}=\frac{n(n+1)}{2}$

Using these formulas, we obtain the number of stones in the pyramid to be $1333300$ . The largest perfect cube less than this number is $110^3=1331000$ . Thus, there are $\boxed{2300}$ stones left over.

Working this problem, I wrote the first few partial sums of the series $1^2 + 3^2 + 5^2 + 7^2 + \cdots$ : $1,\ 10,\ 35,\ 84,\ 165,\ \dots$ This "feels" very much like Pascal's triangle... indeed, it appears that $\underbrace{1^2 + 3^2 + 5^2 + \cdots + (2n-1)^2}_{n\ \text{terms}} = \left(\begin{array}{c} 2n+1 \\ 3 \end{array}\right).$ (Proof: see below.)

The number of stones is therefore $\left(\begin{array}{c} 201 \\ 3 \end{array}\right) = \frac{201\cdot200\cdot199}{3\cdot 2\cdot 1} = 67\cdot 100\cdot 199 = 1\:333\:300,$ which is easily recognized as slightly more than $1\:331\:000 = 110^3$ . The answer is therefore $1\:333\:300 - 1\:331\:000 = \boxed{2300}.$

Proof of the claim made above, by induction:

• Clearly, for just one term we have $1^2 = 1 = \left(\begin{array}{c} 3 \\ 3 \end{array}\right).$

• Suppose the statement is true for $n$ . We will prove it is also true for $n' = n+1$ . Add the $(n+1)$ th term: $(1^2 + 3^2 + 5^2 + \cdots + (2n-1)^2) + (2n+1)^2 \\ = \left(\begin{array}{c} 2n+1 \\ 3 \end{array}\right) + (2n+1)^2 \\ = \frac{(2n+1)(2n)(2n-1)}{6} + (2n+1)^2 \\ = \frac{2n+1}{6}\left( (2n)(2n-1) + 6(2n+1)\right) \\ = \frac{2n+1}{6}(4n^2+10n+6) \\ = \frac{2n+1}{6}(2n+2)(2n+3) \\ = \frac{(2n'+1)(2n')(2n'-1)}{6}\\ = \left(\begin{array}{c} 2n'+1 \\ 3 \end{array}\right).$

Another way to find the sum of the odd squares is to simply add the sum of all of the squares up to 199, and then subtract the even squares up to 198, which can be recognized as 4 times the sum of all of the squares up to 99.