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1992 AIME Problems/Problem 8

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Contents

1 Problem
2 Solution
3 Alternate Solution
4 See also

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

Since the second differences are all $1$ and $a_{19}=a_{92}^{}=0$ , $a_n$ can be expressed explicitly by the quadratic: $a_n=\frac{1}{2!}(n-19)(n-92)$ .

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

Alternate Solution

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, $a_1=\boxed{819}$ .

See also

1992 AIME (Problems • Resources)
Preceded by Problem 7	Followed by Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

Retrieved from "http://www.artofproblemsolving.com/Wiki/index.php/1992_AIME_Problems/Problem_8"

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