2022 AMC 10B Problems/Problem 9
Contents
- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Solution 3 (Induction)
- 5 Solution 4
- 6 Solution 5 (Combinatorics)
- 7 Solution 6 (Desperate)
- 8 Solution 7 (Partial Fractions)
- 9 Solution 8 (Taylor Series)
- 10 Solution 9 (really, really fast cheese)
- 11 Video Solution
- 12 Video Solution by Interstigation
- 13 Video Solution by paixiao
- 14 See Also
Problem
The sum can be expressed as , where and are positive integers. What is ?
Solution 1
Note that , and therefore this sum is a telescoping sum, which is equivalent to . Our answer is .
~mathboy100
Solution 2
We add to the original expression This sum clearly telescopes, thus we end up with . Thus the original expression is equal to , and .
~not_slay (+ minor LaTeX edit ~TaeKim)
Solution 3 (Induction)
By looking for a pattern, we see that and , so we can conclude by engineer's induction that the sum in the problem is equal to , for an answer of . This can be proven with actual induction as well; we have already established base cases, so now assume that for . For we get , completing the proof. ~eibc
Solution 4
Let
Note that Therefore, the answer is
~lopkiloinm
Solution 5 (Combinatorics)
Let's examine a tuple containing distinct integers. We want to find the probability of the tuple being unsorted.
Suppose that we are looking at the first two items in our tuple. The probability of the first element being greater than the second element is .
When we are looking at the first three items in our tuple, the probability of the second element being greater than the third element and the first element less than or equal to the second element is .
Similarly, when we are looking at the first four items in our tuple, the probability of the third element being greater than the fourth element, the second element less than or equal to the third element, and the first element less than or equal to the second element is .
More specifically,
This series ends up being the probability of having the list unsorted and that is of course
~lopkiloinm
Solution 6 (Desperate)
Because the fractions get smaller, it is obvious that the answer is less than , so we can safely assume that (this can also be guessed by intuition using similar math problems). Looking at the answer choices, . Because the last term consists of (and the year is ) we can guess that . Adding them yields .
~iluvme and andy_lee
Solution 7 (Partial Fractions)
Knowing that the answer will be in the form , we can guess that the sum telescopes. Using partial fractions, we can hope to rewrite as . Setting these equal and multiplying by , we get . Since is the only term with with degree , we can conclude that . This means that . Substituting, we find that . This sum clearly telescopes and we obtain . This means that our desired answer is
~kn07
Solution 8 (Taylor Series)
We calculate the Taylor series error to be and this error happens times so it is which is
Solution 9 (really, really fast cheese)
Note that , so we guess and . We get
Video Solution
by Ismail.maths https://www.youtube.com/watch?v=lt34QscjTf4&list=PLmpPPbOoDfgj5BlPtEAGcB7BR_UA5FgFj
- Whiz
Video Solution by Interstigation
https://youtu.be/_KNR0JV5rdI?t=1102
Video Solution by paixiao
https://www.youtube.com/watch?v=MRwZ4F7Scog
See Also
2022 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
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 10 Problems and Solutions |
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