Difference between revisions of "2022 AMC 10B Problems/Problem 9"
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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. | 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. | ||
~eibc | ~eibc | ||
| + | ==Solution 4== | ||
| + | <cmath>\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\dots+\frac{2021}{2022!}</cmath> | ||
== Video Solution == | == Video Solution == | ||
Revision as of 02:27, 22 November 2022
Contents
Problem
The sum
![\[\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\dots+\frac{2021}{2022!}\]](http://latex.artofproblemsolving.com/9/e/3/9e3cf91ad68f1674a89bb502c2853024eb12b445.png)
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 (+ 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
Video Solution
- Whiz
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
