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

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 $\begin{align*} 91\left(17a-\frac{17\cdot18}{2}\right)&=18\left(90a-\frac{90\cdot91}{2}\right)\\ 1547a-\frac{(18\cdot91)\cdot17}{2}&=1620-\frac{(18\cdot91)\cdot90}{2}\\ 73a&=\frac{18\cdot91}{2}\cdot(90-17=73)\\ a=\frac{18\cdot91}{2}=9\cdot91=\boxed{819}. \end{align*}$

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 == 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 thet 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>.
-== Solution ==
+== Solution 1 (uses calculus)==
-{{solution}}
+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>
+This inspires us to guess a quadratic with leading coefficient 1/2 as the solution;
+<cmath> a_{n} = \frac{1}{2}(n-19)(n-92) </cmath>
+as we must have roots at <math>n = 19</math> and <math>n = 92</math>.
-== See also ==
+Thus, <math>a_1=\frac{1}{2}(1-19)(1-92)=\boxed{819}</math>.
-* [[1992 AIME Problems/Problem 7 | Previous Problem]]
+== Solution 2 ==
+Let <math>\Delta^1 A=\Delta A</math>, and <math>\Delta^n A=\Delta(\Delta^{(n-1)}A)</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>
+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>.
+== 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}
+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}
+</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>
+We can sum equation <math>(1)</math> from <math>k=1</math> to <math>18</math>, finding
+<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
+<cmath>91(a_1-a_2) - a_1 = 4095.\tag{3}</cmath>
+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>.
+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 <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>
-* [[1992 AIME Problems/Problem 9 | Next Problem]]
+==Solution 6==
+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*}
+a_3&=2b+1-a \\
+a_4&=3b+3-2a\\
+a_5&=4b+6-3a,
+\end{align*}</cmath>
+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
+<cmath>\begin{align*}
+=&18b+\frac{17\cdot18}{2}-17a\\
+=&91b+\frac{90\cdot91}{2}-90a\\
+b=&\frac{17a-\frac{17\cdot18}{2}}{18}=\frac{90a-\frac{90\cdot91}{2}}{91}.
+\end{align*}</cmath>
+Now, cross multiplying, we find that
+<cmath>\begin{align*}
+\left(17a-\frac{17\cdot18}{2}\right)&=18\left(90a-\frac{90\cdot91}{2}\right)\\
+a-\frac{(18\cdot91)\cdot17}{2}&=1620-\frac{(18\cdot91)\cdot90}{2}\\
+a&=\frac{18\cdot91}{2}\cdot(90-17=73)\\
+a=\frac{18\cdot91}{2}=9\cdot91=\boxed{819}.
+\end{align*}</cmath>
-* [[1992 AIME Problems]]
+== See also ==
+{{AIME box|year=1992|num-b=7|num-a=9}}
+{{MAA Notice}}

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