2024 AMC 12A Problems/Problem 21
Contents
[hide]Problem
Suppose that and the sequence satisfies the recurrence relation for all What is the greatest integer less than or equal to
Solution 1
Multiply both sides of the recurrence to find that .
Let . Then the previous relation becomes
We can rewrite this relation for values of until and use telescoping to derive an explicit formula:
Summing the equations yields:
Now we can substitute back into our equation:
Thus the sum becomes
We know that , and we also know that , so the requested sum is equivalent to . All that remains is to calculate , and we know that this value lies between and (see the note below for a proof). Thus,
so
and thus the answer is .
~eevee9406
Note: . It is obvious that the sum is greater than 1 (since it contains as one of its terms).
If you forget this and have to derive this on the exam, here is how:
and it is clear that . ~eevee9406
Solution 2
According to the equation given,
Suppose , , then , where , then we obtain that
Hence,
Notice that,
so
and thus the answer is .
Solution 3 (lazy + quick)
We'll first try to isolate in terms of .
Now, as with many, many of these large summation problems, if we just evaluate the first few values in the series, a pattern should emerge quickly. Here it works out well since our product on the LHS cancels out.
Here it becomes glaringly obvious that .
So, .
We proceed with the same summation strategy as Solution 1 and get our answer of .
- Note: You only have find the answer's units digit from the answer choices; that's for each sum, giving choice B.
~nm1728
Solution 4 (transform)
Set
Solution 5 (transform)
According to the equation given,
The rest continues similar to Solution 1 or 2
See also
2024 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 20 |
Followed by Problem 22 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.