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

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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{189}. \end{align*}

1992 AIME (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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