Difference between revisions of "1992 AIME Problems/Problem 8"
The 76923th (talk | contribs) (→Solution 6) |
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1547a-\frac{(18\cdot91)\cdot17}{2}&=1620-\frac{(18\cdot91)\cdot90}{2}\\ | 1547a-\frac{(18\cdot91)\cdot17}{2}&=1620-\frac{(18\cdot91)\cdot90}{2}\\ | ||
73a&=\frac{18\cdot91}{2}\cdot(90-17=73)\\ | 73a&=\frac{18\cdot91}{2}\cdot(90-17=73)\\ | ||
− | a=\frac{18\cdot91}{2}=9\cdot91=\boxed{ | + | a=\frac{18\cdot91}{2}=9\cdot91=\boxed{819}. |
\end{align*}</cmath> | \end{align*}</cmath> | ||
Latest revision as of 21:36, 3 December 2022
Contents
Problem
For any sequence of real numbers , define
to be the sequence
, whose
term is
. Suppose that all of the terms of the sequence
are
, and that
. Find
.
Solution 1 (uses calculus)
Note that the s are reminiscent of differentiation; from the condition
, we are led to consider the differential equation
This inspires us to guess a quadratic with leading coefficient 1/2 as the solution;
as we must have roots at
and
.
Thus, .
Solution 2
Let , and
.
Note that in every sequence of ,
Then
Since ,
Solving, .
Solution 3
The sequence is the second finite difference sequence, and the first
terms of this sequence can be computed in terms of the original sequence as shown below.
Adding the above equations we find that
We can sum equation from
to
, finding
We can also sum equation from
to
, finding
Finally,
gives
.
Kris17
Solution 4
Since all terms of are 1, we know that
looks like
for some
. This means
looks like
. More specifically,
. Plugging in
, we have the following linear system:
From this, we can easily find that
and
.
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 so the quadratic is
for some constant
Now we use the conditions that the finite difference is
to find
We know
and
and
Therefore applying finite differences once yields the sequence
and then applying finite differences one more time yields
so
Therefore
Solution 6
Let From the conditions, we have
for all
From this, we find that
or, in general,
Note: we can easily prove this by induction. Now, substituting
we find that
Now, cross multiplying, we find that
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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