The solution leads to a rather natural question: is there a formula for summing the first n consecutive perfect squares: 1 + 4 + 9 + . . . + n^2? This is not an easy formula to derive and I didn't ask the students to do so. Rather, I gave them the formula, which is [n(n+1)(2n + 1)]/6 and we first checked our result for the checkerboard problem against actually doing the sum. We then spent some time building a three dimensional model that serves as a visual proof of the formula (for those familiar with mathematical induction, it's relatively easy to prove the formula by induction, but that is probably too advanced right now for the students, and more importantly, that proof doesn't give any insight into how the formula was arrived at in the first place.
Here are some photos of what the students built from linking cubes:
What the students did was to build six identical "staircases" with 1, 4, and 9 steps. Then, they made two complementary shapes, each constructed from three of the staircases, and finally fit the two large structures together to make a rectangular solid. The dimensions of the solid, as you may be able to discern from the various photographs, are 3 x 4 x 7. This gives a volume of 84 for the solid. Since there are six staircases, we divide 84 by 6 to get 14, which is the sum of 1 + 4 + 9, the first three perfect square numbers. And plugging n = 3 into the formula gives exactly 3(3 + 1)(2*3 + 1) = 3*4*7 = 84.
We still have more work to do with this formula and will return to it (and some other approaches to "proving" it) later in the term. A couple of students have asked me to leave the materials so they can try building a solid made from staircases with 4 or even 5 levels when they have time outside of math class.
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