Difference between revisions of "2022 AMC 10B Problems/Problem 9"
Lopkiloinm (talk | contribs) (→Solution 4) |
Lopkiloinm (talk | contribs) (→Solution 4) |
||
Line 25: | Line 25: | ||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
\left(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\dots+\frac{2021}{2022!}\right)+\left(\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\dots+\frac{1}{2022!}\right)&=\left(\frac{1}{1!}+\frac{2}{2!}+\frac{3}{3!}+\dots+\frac{2022}{2022!}\right)\\ | \left(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\dots+\frac{2021}{2022!}\right)+\left(\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\dots+\frac{1}{2022!}\right)&=\left(\frac{1}{1!}+\frac{2}{2!}+\frac{3}{3!}+\dots+\frac{2022}{2022!}\right)\\ | ||
− | &=\left(\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\dots+\frac{1}{2021!}\right) | + | \left(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\dots+\frac{2021}{2022!}\right)+x&=\left(\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\dots+\frac{1}{2021!}\right) |
\end{align*}</cmath> | \end{align*}</cmath> | ||
Revision as of 03:36, 22 November 2022
Contents
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 (+ 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
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.