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Difference between revisions of "1992 AIME Problems/Problem 8"
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Thus, <math>a_1=\frac{1}{2!}(1-19)(1-92)=\boxed{819}</math>.  
 
Thus, <math>a_1=\frac{1}{2!}(1-19)(1-92)=\boxed{819}</math>.  
  
 +
== Alternate Solution ==
 +
Note that in every sequence of <math>a_i</math>, <math>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 + ...</math>
 +
 +
Then <math>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 + ...</math>
 +
 +
Since <math>\Delta a_1 =a_2 -a_1</math>, <math>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}</math>
 +
 +
<math>a_{19}=0=a_1+18(a_2-a_1)+\binom{18}{2}=18a_2-17a_1+153</math>
 +
 +
<math>a_{92}=0=a_1+91(a_2-a_1)+\binom{91}{2}=91a_2-90a_1+4095</math>
 +
 +
Solving, <math>a_1=\boxed{819}</math>
 
== See also ==
 
== See also ==
 
{{AIME box|year=1992|num-b=7|num-a=9}}
 
{{AIME box|year=1992|num-b=7|num-a=9}}

Revision as of 21:23, 15 March 2009

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

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

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 • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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