part (i) of Fig. 3.5 we have the first six rows of Pascal’s triangle, where a hexagon cent...

part (i) of Fig. 3.5 we have the first six rows of Pascal’s triangle, where a hexagon centered at 4 appears in the last three rows. If we consider the six numbers (around 4) at the vertices of this hexagon, we find that the two alternating triples — namely, 3, 1, 10 and 1, 5, 6 — satisfy 3 • 1 • 10 = 30 = 1 • 5 • 6. Part (ii) of the figure contains rows 4 through 7 of Pascal’s triangle. Here we find a hexagon centered at 10, and the alternating triples at the vertices — in this case, 4, 10, 15 and 6, 20, 5 —satisfy 4 • 10 • 15 = 600 = 6 • 20 • 5.

a) Conjecture the general result suggested by these two examples.

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b) Verify the conjecture in part (a).