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Difference between revisions of "1992 AIME Problems/Problem 8"
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== Problem ==
 
== Problem ==
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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>.
+
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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== Solution ==
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== Solution 1 (uses calculus)==
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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>
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This inspires us to guess a quadratic with leading coefficient 1/2 as the solution;
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<cmath> a_{n} = \frac{1}{2}(n-19)(n-92) </cmath>
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as we must have roots at <math>n = 19</math> and <math>n = 92</math>.
  
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Thus, <math>a_1=\frac{1}{2!}(1-19)(1-92)=\boxed{819}</math>.  
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Thus, <math>a_1=\frac{1}{2}(1-19)(1-92)=\boxed{819}</math>.
 +
 
 +
== Solution 2 ==
 +
Let <math>\Delta^1 A=\Delta A</math>, and <math>\Delta^n A=\Delta(\Delta^{(n-1)}A)</math>.
  
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== 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>
 
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>
  
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<math>a_{92}=0=a_1+91(a_2-a_1)+\binom{91}{2}=91a_2-90a_1+4095</math>
 
<math>a_{92}=0=a_1+91(a_2-a_1)+\binom{91}{2}=91a_2-90a_1+4095</math>
  
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Solving, <math>a_1=\boxed{819}</math>
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Solving, <math>a_1=\boxed{819}</math>.
 +
 
 +
== Solution 3 ==
 +
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>\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>
 +
 
 +
Adding the above <math>k-1</math> equations we find that
 +
 
 +
<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>
 +
 
 +
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>
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Finally, <math>18\cdot (3) - 91\cdot(2)</math> gives <math>a_1=\boxed{819}</math>.
 +
 
 +
Kris17
 +
 
 +
== Solution 4 ==
 +
Since all terms of <math>\Delta(\Delta A)</math> are 1, we know that <math>\Delta A</math> looks like <math>(k,k+1,k+2,...)</math> for some <math>k</math>. This means <math>A</math> looks like <math>(a_1,a_1+k,a_1+2k+1,a_1+3k+3,a_1+4k+6,...)</math>. More specifically, <math>A_n=a_1+k(n-1)+\frac{(n-1)(n-2)}{2}</math>. Plugging in <math>a_{19}=a_{92}=0</math>, we have the following linear system: <cmath>a_1+91k=-4095</cmath> <cmath>a_1+18k=-153</cmath> From this, we can easily find that <math>k=-54</math> and <math>a_1=\boxed{819}</math>.
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Solution by Zeroman
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 +
 
 +
 
 +
==Solution 5 ==
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 +
Since the result of two finite differences of some sequence is a constant sequence, we know that sequence is a quadratic. Furthermore, we know that <math>f(19) = f(92) = 0</math> so the quadratic is <math>f(x) = a(x-19)(x-92)</math> for some constant <math>a.</math> Now we use the conditions that the finite difference is <math>1</math> to find <math>a.</math> We know <math>f(19) = 0</math> and <math>f(20) = -72a</math> and <math>f(18) = 74a.</math> Therefore applying finite differences once yields the sequence <math>-74a,-72a</math> and then applying finite differences one more time yields <math>2a</math> so <math>a =\frac{1}{2}.</math> Therefore <math>f(1) = 9 \cdot 91 = \boxed{819}.</math>
 +
 
 +
==Solution 6==
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Let <math>a_1=a,a_2=b.</math> From the conditions, we have <cmath>a_{n-1}+a_{n+1}=2a_n+1,</cmath> for all <math>n>1.</math> From this, we find that
 +
<cmath>\begin{align*}
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a_3&=2b+1-a \\
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a_4&=3b+3-2a\\
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a_5&=4b+6-3a,
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\end{align*}</cmath>
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or, in general, <cmath>a_n=(n-1)b+\frac{(n-2)(n-1)}{2}-(n-2)a.</cmath> Note: we can easily prove this by induction. Now, substituting <math>n=19,92,</math> we find that
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<cmath>\begin{align*}
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0=&18b+\frac{17\cdot18}{2}-17a\\
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0=&91b+\frac{90\cdot91}{2}-90a\\
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b=&\frac{17a-\frac{17\cdot18}{2}}{18}=\frac{90a-\frac{90\cdot91}{2}}{91}.
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\end{align*}</cmath>
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Now, cross multiplying, we find that
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<cmath>\begin{align*}
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91\left(17a-\frac{17\cdot18}{2}\right)&=18\left(90a-\frac{90\cdot91}{2}\right)\\
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1547a-\frac{(18\cdot91)\cdot17}{2}&=1620-\frac{(18\cdot91)\cdot90}{2}\\
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73a&=\frac{18\cdot91}{2}\cdot(90-17=73)\\
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a=\frac{18\cdot91}{2}=9\cdot91=\boxed{819}.
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\end{align*}</cmath>
 +
 
 
== See also ==
 
== See also ==
 
{{AIME box|year=1992|num-b=7|num-a=9}}
 
{{AIME box|year=1992|num-b=7|num-a=9}}
 +
{{MAA Notice}}

Latest revision as of 20:36, 3 December 2022

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 1 (uses calculus)

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

Solution 4

Since all terms of $\Delta(\Delta A)$ are 1, we know that $\Delta A$ looks like $(k,k+1,k+2,...)$ for some $k$. This means $A$ looks like $(a_1,a_1+k,a_1+2k+1,a_1+3k+3,a_1+4k+6,...)$. More specifically, $A_n=a_1+k(n-1)+\frac{(n-1)(n-2)}{2}$. Plugging in $a_{19}=a_{92}=0$, we have the following linear system: \[a_1+91k=-4095\] \[a_1+18k=-153\] From this, we can easily find that $k=-54$ and $a_1=\boxed{819}$. Solution by Zeroman


Solution 5

Since the result of two finite differences of some sequence is a constant sequence, we know that sequence is a quadratic. Furthermore, we know that $f(19) = f(92) = 0$ so the quadratic is $f(x) = a(x-19)(x-92)$ for some constant $a.$ Now we use the conditions that the finite difference is $1$ to find $a.$ We know $f(19) = 0$ and $f(20) = -72a$ and $f(18) = 74a.$ Therefore applying finite differences once yields the sequence $-74a,-72a$ and then applying finite differences one more time yields $2a$ so $a =\frac{1}{2}.$ Therefore $f(1) = 9 \cdot 91 = \boxed{819}.$

Solution 6

Let $a_1=a,a_2=b.$ From the conditions, we have \[a_{n-1}+a_{n+1}=2a_n+1,\] for all $n>1.$ From this, we find that \begin{align*} a_3&=2b+1-a \\ a_4&=3b+3-2a\\ a_5&=4b+6-3a, \end{align*} or, in general, \[a_n=(n-1)b+\frac{(n-2)(n-1)}{2}-(n-2)a.\] Note: we can easily prove this by induction. Now, substituting $n=19,92,$ we find that \begin{align*} 0=&18b+\frac{17\cdot18}{2}-17a\\ 0=&91b+\frac{90\cdot91}{2}-90a\\ b=&\frac{17a-\frac{17\cdot18}{2}}{18}=\frac{90a-\frac{90\cdot91}{2}}{91}. \end{align*} Now, cross multiplying, we find that \begin{align*} 91\left(17a-\frac{17\cdot18}{2}\right)&=18\left(90a-\frac{90\cdot91}{2}\right)\\ 1547a-\frac{(18\cdot91)\cdot17}{2}&=1620-\frac{(18\cdot91)\cdot90}{2}\\ 73a&=\frac{18\cdot91}{2}\cdot(90-17=73)\\ a=\frac{18\cdot91}{2}=9\cdot91=\boxed{819}. \end{align*}

See also

1992 AIME (Problems • Answer Key • Resources)
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