Difference between revisions of "2022 AMC 10B Problems/Problem 9"
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+ | ==Solution 3 (Induction)== | ||
+ | By looking for a pattern, we see that <math>\tfrac{1}{2!} = 1 - \tfrac{1}{2!}</math> and <math>\tfrac{1}{2!} + \tfrac{2}{3!} = \tfrac{5}{6} = 1 - \tfrac{1}{3!}</math>, so we can conclude by engineer's induction that the sum in the problem is equal to <math>1 - \tfrac{1}{2022!}</math>, for an answer of <math>\boxed{\textbf{(D)}\ 2023}</math>. This can be proven with actual induction as well; we have already established <math>2</math> base cases, so now assume that <math>\tfrac{1}{2!} + \tfrac{2}{3!} + \cdots \tfrac{n-1}{n!} = 1 - \tfrac{1}{n!}</math> for <math>n = k</math>. For <math>n = k + 1</math> we get <math>\tfrac{1}{2!} + \tfrac{2}{3!} + \cdots \tfrac{n-1}{n!} + \tfrac{n}{(n+1)!} = 1 - \tfrac{1}{n!} + \tfrac{n}{(n+1)!} = 1 - \tfrac{n+1}{(n+1)!} + \tfrac{n}{(n+1)!} = 1 - \tfrac{1}{(n+1)!}</math>, completing the proof. | ||
== See Also == | == See Also == | ||
{{AMC10 box|year=2022|ab=B|num-b=8|num-a=10}} | {{AMC10 box|year=2022|ab=B|num-b=8|num-a=10}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 18:19, 18 November 2022
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 have from canceling a 2022 from . This sum clearly telescopes, thus we end up with . Thus the original equation is equal to , and . .
~not_slay
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.
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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