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Difference between revisions of "1992 AIME Problems/Problem 8"
(→‎Alternate Solution)
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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 ==
+
== Solution 2 ==
 
Let <math>\Delta^1 A=\Delta A</math>, and <math>\Delta^n A=\Delta(\Delta^{(n-1)}A)</math>.
 
Let <math>\Delta^1 A=\Delta A</math>, and <math>\Delta^n A=\Delta(\Delta^{(n-1)}A)</math>.
  
Line 21: Line 21:
  
 
Solving, <math>a_1=\boxed{819}</math>.
 
Solving, <math>a_1=\boxed{819}</math>.
 +
 +
== Solution 3 ==
 +
Write out and add first <math>k-1</math> terms of the second finite difference sequence:
 +
 +
<math>a_3+a_1-2*a_2=1</math>
 +
 +
<math>a_4+a_2-2*a_3=1</math>
 +
 +
…
 +
 +
…
 +
 +
…
 +
 +
<math>a_k + a_{k-2} - 2*a_{k-1} = 1</math>
 +
 +
<math>a_{k+1} + a_{k-1} - 2*a_k = 1</math>
 +
 +
Adding the above <math>k-1</math> equations we get:
 +
 +
<math>\boxed{(a_{k+1} - a_k) = k-1 + (a_2-a_1)} --- (1)</math>
 +
 +
Now taking sum <math>k = 1</math> to <math>18</math> in equation <math>(1)</math> we get:
 +
<math>18*(a_1-a_2) - a_1 = 153 --- (2)</math>
 +
 +
Now taking sum <math>k = 1</math> to <math>91</math> in equation <math>(1)</math> we get:
 +
<math>91*(a_1-a_2) - a_1 = 4095 --- (3)</math>
 +
 +
<math>18* (3) - 91*(2)</math> gives <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 14:15, 10 December 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, $a_1=\frac{1}{2!}(1-19)(1-92)=\boxed{819}$.

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 $a_1=\boxed{819}$.

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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