Problem:
Compute (31)(30)(29)(28)+1.
Solution:
The data
4⋅3⋅2⋅1+15⋅4⋅3⋅2+16⋅5⋅4⋅3+1=52=(3⋅2−1)2=112=(4⋅3−1)2=192=(5⋅4−1)2
suggest that (k+1)(k)(k−1)(k−2)+1=[k(k−1)−1]2=[(k2−k)−1]2. The calculations
(k+1)(k)(k−1)(k−2)+1=[(k+1)(k−2)][k(k−1)]+1=(k2−k−2)(k2−k)+1=(k2−k)2−2(k2−k)+1=[(k2−k)−1]2
show that this is true. Thus (31)(30)(29)(28)+1=302−30−1=869.
The problems on this page are the property of the MAA's American Mathematics Competitions