In 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 ˆ™ 1 ˆ™ 10 = 30 = 1 ˆ™ 5 ˆ™ 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 ˆ™ 10 ˆ™ 15 = 600 = 6 ˆ™ 20 ˆ™ 5.
(a) Conjecture the general result suggested by these two examples.
(b) Verify the conjecture in part (a).

Figure 3.5

1 4 6 4 1 1 5 10 10 5/1 15 10 10 5 1 1 615 20/15 6 1