Problem:
For any sequence of real numbers A=(a1,a2,a3,…), define △A to be the sequence (a2−a1,a3−a2,a4−a3,…), whose nth term is an+1−an. Suppose that all of the terms of the sequence △(△A) are 1, and that a19=a92=0. Find a1.
Solution:
Suppose that the first term of the sequence △A is d. Then the sequence △A is (d,d+1,d+2,…) with nth term given by d+(n−1). Hence the sequence A is
(a1,a1+d,a1+d+(d+1),a1+d+(d+1)+(d+2),…)
with nth term given by
an=a1+(n−1)d+21(n−1)(n−2)
This shows that an is a quadratic polynomial in n with leading coefficient 21. Since a19=a92=0, we must have
an=21(n−19)(n−92),
so a1=21(1−19)(1−92)=819.
The problems on this page are the property of the MAA's American Mathematics Competitions