# 1992 AIME Problems/Problem 8

## Problem

For any sequence of real numbers $A=(a_1,a_2,a_3,\ldots)$, define $\Delta A^{}_{}$ to be the sequence $(a_2-a_1,a_3-a_2,a_4-a_3,\ldots)$, whose $n^{\mbox{th}}_{}$ term is $a_{n+1}-a_n^{}$. Suppose that all of the terms of the sequence $\Delta(\Delta A^{}_{})$ are $1^{}_{}$, and that $a_{19}=a_{92}^{}=0$. Find $a_1^{}$.

## Solution 1 (uses calculus)

Note that the $\Delta$s are reminiscent of differentiation; from the condition $\Delta(\Delta{A}) = 1$, we are led to consider the differential equation $$\frac{d^2 A}{dn^2} = 1$$ This inspires us to guess a quadratic with leading coefficient 1/2 as the solution; $$a_{n} = \frac{1}{2}(n-19)(n-92)$$ as we must have roots at $n = 19$ and $n = 92$.

Thus, $a_1=\frac{1}{2}(1-19)(1-92)=\boxed{819}$.

## Solution 2

Let $\Delta^1 A=\Delta A$, and $\Delta^n A=\Delta(\Delta^{(n-1)}A)$.

Note that in every sequence of $a_i$, $a_n=\binom{n-1}{1}\Delta a_n + \binom{n-1}{2}\Delta^2 a_n +\binom{n-1}{3}\Delta^3 a_n + ...$

Then $a_n=a_1 +\binom{n-1}{1}\Delta a_1 +\binom{n-1}{2}\Delta^2 a_1 +\binom{n-1}{3}\Delta^3 a_1 + ...$

Since $\Delta a_1 =a_2 -a_1$, $a_n=a_1 +\binom{n-1}{1}(a_2-a_1) +\binom{n-1}{2}\cdot 1=a_1 +n(a_2-a_1) +\binom{n-1}{2}$ $a_{19}=0=a_1+18(a_2-a_1)+\binom{18}{2}=18a_2-17a_1+153$ $a_{92}=0=a_1+91(a_2-a_1)+\binom{91}{2}=91a_2-90a_1+4095$

Solving, $a_1=\boxed{819}$.

## Solution 3

The sequence $\Delta(\Delta A)$ is the second finite difference sequence, and the first $k-1$ terms of this sequence can be computed in terms of the original sequence as shown below. $\begin{array}{rcl} a_3+a_1-2a_2&=&1\\ a_4+a_2-2a_3&=&1\\ &\vdots\\ a_k + a_{k-2} - 2a_{k-1} &= &1\\ a_{k+1} + a_{k-1} - 2a_k &=& 1.\\ \end{array}$

Adding the above $k-1$ equations we find that $$(a_{k+1} - a_k) = k-1 + (a_2-a_1).\tag{1}$$

We can sum equation $(1)$ from $k=1$ to $18$, finding $$18(a_1-a_2) - a_1 = 153.\tag{2}$$

We can also sum equation $(1)$ from $k=1$ to $91$, finding $$91(a_1-a_2) - a_1 = 4095.\tag{3}$$ Finally, $18\cdot (3) - 91\cdot(2)$ gives $a_1=\boxed{819}$.

Kris17

## Solution 4

Since all terms of $\Delta(\Delta A)$ are 1, we know that $\Delta A$ looks like $(k,k+1,k+2,...)$ for some $k$. This means $A$ looks like $(a_1,a_1+k,a_1+2k+1,a_1+3k+3,a_1+4k+6,...)$. More specifically, $A_n=a_1+k(n-1)+\frac{(n-1)(n-2)}{2}$. Plugging in $a_{19}=a_{92}=0$, we have the following linear system: $$a_1+91k=-4095$$ $$a_1+18k=-153$$ From this, we can easily find that $k=-54$ and $a_1=\boxed{819}$. Solution by Zeroman

## Solution 5

Since the result of two finite differences of some sequence is a constant sequence, we know that sequence is a quadratic. Furthermore, we know that $f(19) = f(92) = 0$ so the quadratic is $f(x) = a(x-19)(x-92)$ for some constant $a.$ Now we use the conditions that the finite difference is $1$ to find $a.$ We know $f(19) = 0$ and $f(20) = -72a$ and $f(18) = 74a.$ Therefore applying finite differences once yields the sequence $-74a,-72a$ and then applying finite differences one more time yields $2a$ so $a =\frac{1}{2}.$ Therefore $f(1) = 9 \cdot 91 = \boxed{819}.$

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 