Difference between revisions of "1992 AIME Problems/Problem 8"

Revision as of 10:54, 30 July 2013

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 $n^\mbox{th}_{}$ (Error compiling LaTeX. Unknown error_msg) 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

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; $a_{n} = \frac{1}{2}(n-19)(n-94)$ 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

Write out and add first $k-1$ terms of the second finite difference sequence:

$a_3+a_1-2*a_2=1$

$a_4+a_2-2*a_3=1$

…

$a_k + a_{k-2} - 2*a_{k-1} = 1$

$a_{k+1} + a_{k-1} - 2*a_k = 1$

Adding the above $k-1$ equations we get:

$\boxed{(a_{k+1} - a_k) = k-1 + (a_2-a_1)} --- (1)$

Now taking sum $k = 1$ to $18$ in equation $(1)$ we get: $18*(a_1-a_2) - a_1 = 153 --- (2)$

Now taking sum $k = 1$ to $91$ in equation $(1)$ we get: $91*(a_1-a_2) - a_1 = 4095 --- (3)$

$18* (3) - 91*(2)$ gives $a_1=\boxed{819}$ .

@@ Line 3: / Line 3: @@
 == Solution ==
-Since the second differences are all <math>1</math> and <math>a_{19}=a_{92}^{}=0</math>, <math>a_n</math> can be expressed explicitly by the quadratic: <math>a_n=\frac{1}{2!}(n-19)(n-92)</math>.
+Note that the <math>\Delta</math>s are reminiscent of differentiation; from the condition <math>\Delta(\Delta{A}) = 1</math>, we are led to consider the differential equation
+<cmath> \frac{d^2 A}{dn^2} = 1 </cmath>
-Thus, <math>a_1=\frac{1}{2!}(1-19)(1-92)=\boxed{819}</math>.
+This inspires us to guess a quadratic with leading coefficient 1/2 as the solution;
+<cmath> a_{n} = \frac{1}{2}(n-19)(n-94) </cmath>
+Thus, <math>a_1=\frac{1}{2}(1-19)(1-92)=\boxed{819}</math>.
 == Solution 2 ==

1992 AIME (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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