How I Discovered the Sum of Squares Formula Over Dinner
It started with a casual dinner at a restaurant with my parents and sister. As always, I brought with me my notebook, two pens, and a brain that refuses to stop thinking. A few moments after sitting down, a curious question struck me and followed me home. I worked on it at the dinner table, in the car, and in my room well into that night. What’s the sum of the first $N$ perfect squares? I’d never seen the formula before, let alone any derivation. Most textbook derivations rely on advanced algebraic identities or clever manipulations I didn’t know at the time. But that day, I found a way that was intuitive and deeply satisfying. What follows is one of my proudest little mathematical discoveries from high school.
Problem at hand:
Can we find a closed-form formula for the sum of the first $N$ perfect squares?
Note that I will not focus on presenting a formal mathematical proof here, rather I’ll simplify my derivation as much as possible, to ensure any audience with a basic mathematical acumen can understand it.
To start, recall (or, as I did, discover after about about 20 minutes of playing around) that the sum of the first $N$ odd natural numbers is $N^2$. If you’ve never heard of this before, it’s a good idea to verify it yourself, since the entire derivation stems from this fact.
For instance:
\[1^2+2^2+3^2=1+4+9=(1)+(1+3)+(1+3+5)\]We note that the number $1$ appears in the sum three times, $3$ appears twice, and $5$ appears only once. The pattern is obvious here because each perfect square is simply the previous perfect square plus the next odd natural number.
Thus, we deduce that:
\[\begin{align*}\sum_{i=1}^{N} i^2 = \sum_{i=1}^{N} (2i - 1)(N + 1 - i )\\\implies \sum_{i=1}^{N} i^2 = \left( \sum_{i=1}^{N} (2i - 1) \right) N - \sum_{i=1}^{N-1} i(2i + 1)\end{align*}\]We know that $\left( \sum_{i=1}^{N} (2i - 1) \right) = N^2$ because the sum of the first $N$ odd natural numbers is $N^2$ (a fact we discussed earlier in this post).
Then,
\[\begin{align*} \sum_{i=1}^{N} i^2 &= N^3 - \sum_{i=1}^{N-1} i(2i + 1) \\ &=N^3-2\sum_{i=1}^{N} i^2 - \sum_{i=1}^{N} i + 2N^2 + N\\\implies 3\sum_{i=1}^{N} i^2 &=N^3-\sum_{i=1}^{N} i+N(2N+1)\end{align*}\]Recall that $\sum_{i=1}^{N} i = \frac{N(N+1)}{2}$:
\[\begin{align*} 3\sum_{i=1}^{N} i^2 &= N^3 - \frac{N(N+1)}{2} + N(2N+1) \\ &= N^3 + 2N^2 + N - \frac{N(N+1)}{2} \\ &= \frac{2N^3 + 4N^2 + 2N - N(N+1)}{2} \\ &= \frac{2N^3 + 4N^2 + 2N - N^2 - N}{2} \\ &= \frac{2N^3 + 3N^2 + N}{2} \end{align*}\]Dividing both sides by $3$ gives us the final result:
\[\sum_{i=1}^{N} i^2 = \frac{2N^3 + 3N^2 + N}{6} = \frac{N(N+1)(2N+1)}{6}\]And there it is, the beautiful closed-form formula for the sum of the first $N$ perfect squares, derived without ever having seen it before. Just a notebook, two pens, and an unusually persistent brain over dinner :)
I’d recommend every high schooler to give this a try, and if you know someone who would benefit from this exercise, tell them to derive the formula without seeing it. I firmly believe anyone can solve anything (without guidance) with enough time and curiosity. Perhaps you can discover a more intuitive method to solve this problem :D