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

(Solution 3)
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== Problem ==
 
== Problem ==
For any sequence of real numbers <math>A=(a_1,a_2,a_3,\ldots)</math>, define <math>\Delta A^{}_{}</math> to be the sequence <math>(a_2-a_1,a_3-a_2,a_4-a_3,\ldots)</math>, whose <math>n^\mbox{th}_{}</math> term is <math>a_{n+1}-a_n^{}</math>. Suppose that all of the terms of the sequence <math>\Delta(\Delta A^{}_{})</math> are <math>1^{}_{}</math>, and that <math>a_{19}=a_{92}^{}=0</math>. Find <math>a_1^{}</math>.
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For any sequence of real numbers <math>A=(a_1,a_2,a_3,\ldots)</math>, define <math>\Delta A^{}_{}</math> to be the sequence <math>(a_2-a_1,a_3-a_2,a_4-a_3,\ldots)</math>, whose <math>n^{\mbox{th}}_{}</math> term is <math>a_{n+1}-a_n^{}</math>. Suppose that all of the terms of the sequence <math>\Delta(\Delta A^{}_{})</math> are <math>1^{}_{}</math>, and that <math>a_{19}=a_{92}^{}=0</math>. Find <math>a_1^{}</math>.
  
 
== Solution ==
 
== Solution ==
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== Solution 3 ==
 
== Solution 3 ==
Write out and add first <math>k-1</math> terms of the second finite difference sequence:
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The sequence <math>\Delta(\Delta A)</math> is the second finite difference sequence, and the first <math>k-1</math> terms of this sequence can be computed in terms of the original sequence as shown below.
  
<math>a_3+a_1-2*a_2=1</math>
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<math>\begin{array}{rcl}
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a_3+a_1-2a_2&=&1\\
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a_4+a_2-2a_3&=&1\\
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&\vdots\\
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a_k + a_{k-2} - 2a_{k-1} &= &1\\
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a_{k+1} + a_{k-1} - 2a_k &=& 1.\\
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\end{array}
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</math>
  
<math>a_4+a_2-2*a_3=1</math>
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Adding the above <math>k-1</math> equations we find that
  
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<cmath>(a_{k+1} - a_k) = k-1 + (a_2-a_1).\tag{1}</cmath>
  
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We can sum equation <math>(1)</math> from <math>k=1</math> to <math>18</math>, finding
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<cmath>18(a_1-a_2) - a_1 = 153.\tag{2} </cmath>
  
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We can also sum equation <math>(1)</math> from <math>k=1</math> to <math>91</math>, finding
 
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<cmath>91(a_1-a_2) - a_1 = 4095.\tag{3}</cmath>
<math>a_k + a_{k-2} - 2*a_{k-1} = 1</math>
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Finally, <math>18\cdot (3) - 91\cdot(2)</math> gives <math>a_1=\boxed{819}</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>.
 
  
 
Kris17
 
Kris17

Revision as of 00:39, 7 March 2015

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

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

See also

1992 AIME (ProblemsAnswer KeyResources)
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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