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