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k a May Highlights and 2025 AoPS Online Class Information
jlacosta   0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.

Are you interested in working towards MATHCOUNTS and don’t know where to start? We have you covered! If you have taken Prealgebra, then you are ready for MATHCOUNTS/AMC 8 Basics. Already aiming for State or National MATHCOUNTS and harder AMC 8 problems? Then our MATHCOUNTS/AMC 8 Advanced course is for you.

Summer camps are starting next month at the Virtual Campus in math and language arts that are 2 - to 4 - weeks in duration. Spaces are still available - don’t miss your chance to have an enriching summer experience. There are middle and high school competition math camps as well as Math Beasts camps that review key topics coupled with fun explorations covering areas such as graph theory (Math Beasts Camp 6), cryptography (Math Beasts Camp 7-8), and topology (Math Beasts Camp 8-9)!

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[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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0 replies
jlacosta
May 1, 2025
0 replies
k i Adding contests to the Contest Collections
dcouchman   1
N Apr 5, 2023 by v_Enhance
Want to help AoPS remain a valuable Olympiad resource? Help us add contests to AoPS's Contest Collections.

Find instructions and a list of contests to add here: https://artofproblemsolving.com/community/c40244h1064480_contests_to_add
1 reply
dcouchman
Sep 9, 2019
v_Enhance
Apr 5, 2023
k i Zero tolerance
ZetaX   49
N May 4, 2019 by NoDealsHere
Source: Use your common sense! (enough is enough)
Some users don't want to learn, some other simply ignore advises.
But please follow the following guideline:


To make it short: ALWAYS USE YOUR COMMON SENSE IF POSTING!
If you don't have common sense, don't post.


More specifically:

For new threads:


a) Good, meaningful title:
The title has to say what the problem is about in best way possible.
If that title occured already, it's definitely bad. And contest names aren't good either.
That's in fact a requirement for being able to search old problems.

Examples:
Bad titles:
- "Hard"/"Medium"/"Easy" (if you find it so cool how hard/easy it is, tell it in the post and use a title that tells us the problem)
- "Number Theory" (hey guy, guess why this forum's named that way¿ and is it the only such problem on earth¿)
- "Fibonacci" (there are millions of Fibonacci problems out there, all posted and named the same...)
- "Chinese TST 2003" (does this say anything about the problem¿)
Good titles:
- "On divisors of a³+2b³+4c³-6abc"
- "Number of solutions to x²+y²=6z²"
- "Fibonacci numbers are never squares"


b) Use search function:
Before posting a "new" problem spend at least two, better five, minutes to look if this problem was posted before. If it was, don't repost it. If you have anything important to say on topic, post it in one of the older threads.
If the thread is locked cause of this, use search function.

Update (by Amir Hossein). The best way to search for two keywords in AoPS is to input
[code]+"first keyword" +"second keyword"[/code]
so that any post containing both strings "first word" and "second form".


c) Good problem statement:
Some recent really bad post was:
[quote]$lim_{n\to 1}^{+\infty}\frac{1}{n}-lnn$[/quote]
It contains no question and no answer.
If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.


For answers to already existing threads:


d) Of any interest and with content:
Don't post things that are more trivial than completely obvious. For example, if the question is to solve $x^{3}+y^{3}=z^{3}$, do not answer with "$x=y=z=0$ is a solution" only. Either you post any kind of proof or at least something unexpected (like "$x=1337, y=481, z=42$ is the smallest solution). Someone that does not see that $x=y=z=0$ is a solution of the above without your post is completely wrong here, this is an IMO-level forum.
Similar, posting "I have solved this problem" but not posting anything else is not welcome; it even looks that you just want to show off what a genius you are.

e) Well written and checked answers:
Like c) for new threads, check your solutions at least twice for mistakes. And after sending, read it again and use the Edit-Button if necessary to correct errors.



To repeat it: ALWAYS USE YOUR COMMON SENSE IF POSTING!


Everything definitely out of range of common sense will be locked or deleted (exept for new users having less than about 42 posts, they are newbies and need/get some time to learn).

The above rules will be applied from next monday (5. march of 2007).
Feel free to discuss on this here.
49 replies
ZetaX
Feb 27, 2007
NoDealsHere
May 4, 2019
Cyclic equality implies equal sum of squares
blackbluecar   33
N 23 minutes ago by Vivaandax
Source: 2021 Iberoamerican Mathematical Olympiad, P4
Let $a,b,c,x,y,z$ be real numbers such that

\[ a^2+x^2=b^2+y^2=c^2+z^2=(a+b)^2+(x+y)^2=(b+c)^2+(y+z)^2=(c+a)^2+(z+x)^2 \]
Show that $a^2+b^2+c^2=x^2+y^2+z^2$.
33 replies
blackbluecar
Oct 21, 2021
Vivaandax
23 minutes ago
3-var inequality
sqing   3
N 36 minutes ago by ytChen
Source: Own
Let $ a,b>0 $ and $\frac{1}{a^2+3}+ \frac{1}{b^2+ 3} \leq \frac{1}{2} . $ Prove that
$$a^2+ab+b^2\geq 3$$$$a^2-ab+b^2 \geq 1 $$Let $ a,b>0 $ and $\frac{1}{a^3+3}+ \frac{1}{b^3+ 3}\leq \frac{1}{2} . $ Prove that
$$a^3+ab+b^3 \geq 3$$$$ a^3-ab+b^3\geq 1 $$
3 replies
sqing
May 7, 2025
ytChen
36 minutes ago
An I for an I
Eyed   67
N an hour ago by AR17296174
Source: 2020 ISL G8
Let $ABC$ be a triangle with incenter $I$ and circumcircle $\Gamma$. Circles $\omega_{B}$ passing through $B$ and $\omega_{C}$ passing through $C$ are tangent at $I$. Let $\omega_{B}$ meet minor arc $AB$ of $\Gamma$ at $P$ and $AB$ at $M\neq B$, and let $\omega_{C}$ meet minor arc $AC$ of $\Gamma$ at $Q$ and $AC$ at $N\neq C$. Rays $PM$ and $QN$ meet at $X$. Let $Y$ be a point such that $YB$ is tangent to $\omega_{B}$ and $YC$ is tangent to $\omega_{C}$.

Show that $A,X,Y$ are collinear.
67 replies
Eyed
Jul 20, 2021
AR17296174
an hour ago
IMO 2018 Problem 1
juckter   169
N 2 hours ago by Thelink_20
Let $\Gamma$ be the circumcircle of acute triangle $ABC$. Points $D$ and $E$ are on segments $AB$ and $AC$ respectively such that $AD = AE$. The perpendicular bisectors of $BD$ and $CE$ intersect minor arcs $AB$ and $AC$ of $\Gamma$ at points $F$ and $G$ respectively. Prove that lines $DE$ and $FG$ are either parallel or they are the same line.

Proposed by Silouanos Brazitikos, Evangelos Psychas and Michael Sarantis, Greece
169 replies
juckter
Jul 9, 2018
Thelink_20
2 hours ago
Urgent. Need them quick
sealight2107   2
N 2 hours ago by Bergo1305
With $a,b,c>1$ and $a+b+c=2abc$. Prove that:
$\sqrt[3]{ab-1}+\sqrt[3]{bc-1}+\sqrt[3]{ca-1} \le \sqrt[3]{(a+b+c)^2}$
2 replies
sealight2107
Yesterday at 4:58 PM
Bergo1305
2 hours ago
Croatian mathematical olympiad, day 1, problem 2
Matematika   6
N 2 hours ago by Cqy00000000
There were finitely many persons at a party among whom some were friends. Among any $4$ of them there were either $3$ who were all friends among each other or $3$ who weren't friend with each other. Prove that you can separate all the people at the party in two groups in such a way that in the first group everyone is friends with each other and that all the people in the second group are not friends to anyone else in second group. (Friendship is a mutual relation).
6 replies
Matematika
Apr 10, 2011
Cqy00000000
2 hours ago
Game
Pascual2005   27
N 3 hours ago by HamstPan38825
Source: Colombia TST, IMO ShortList 2004, combinatorics problem 5
$A$ and $B$ play a game, given an integer $N$, $A$ writes down $1$ first, then every player sees the last number written and if it is $n$ then in his turn he writes $n+1$ or $2n$, but his number cannot be bigger than $N$. The player who writes $N$ wins. For which values of $N$ does $B$ win?

Proposed by A. Slinko & S. Marshall, New Zealand
27 replies
Pascual2005
Jun 7, 2005
HamstPan38825
3 hours ago
Lines concur on bisector of BAC
Invertibility   2
N 5 hours ago by NO_SQUARES
Source: Slovenia 2025 TST 3 P2
Let $\Omega$ be the circumcircle of a scalene triangle $ABC$. Let $\omega$ be a circle internally tangent to $\Omega$ in $A$. Tangents from $B$ touch $\omega$ in $P$ and $Q$, such that $P$ lies in the interior of $\triangle{}ABC$. Similarly, tangents from $C$ touch $\omega$ in $R$ and $S$, such that $R$ lies in the interior of $\triangle{}ABC$.

Prove that $PS$ and $QR$ concur on the bisector of $\angle{}BAC$.
2 replies
Invertibility
5 hours ago
NO_SQUARES
5 hours ago
Why is the old one deleted?
EeEeRUT   16
N 5 hours ago by ravengsd
Source: EGMO 2025 P1
For a positive integer $N$, let $c_1 < c_2 < \cdots < c_m$ be all positive integers smaller than $N$ that are coprime to $N$. Find all $N \geqslant 3$ such that $$\gcd( N, c_i + c_{i+1}) \neq 1$$for all $1 \leqslant i \leqslant m-1$

Here $\gcd(a, b)$ is the largest positive integer that divides both $a$ and $b$. Integers $a$ and $b$ are coprime if $\gcd(a, b) = 1$.

Proposed by Paulius Aleknavičius, Lithuania
16 replies
EeEeRUT
Apr 16, 2025
ravengsd
5 hours ago
angle chasing with 2 midpoints, equal angles given and wanted
parmenides51   5
N 6 hours ago by breloje17fr
Source: Ukrainian Geometry Olympiad 2017, IX p1, X p1, XI p1
In the triangle $ABC$, ${{A}_{1}}$ and ${{C}_{1}} $ are the midpoints of sides $BC $ and $AB$ respectively. Point $P$ lies inside the triangle. Let $\angle BP {{C}_{1}} = \angle PCA$. Prove that $\angle BP {{A}_{1}} = \angle PAC $.
5 replies
parmenides51
Dec 11, 2018
breloje17fr
6 hours ago
Problem 4 of Finals
GeorgeRP   2
N Yesterday at 7:00 PM by Assassino9931
Source: XIII International Festival of Young Mathematicians Sozopol 2024, Theme for 10-12 grade
The diagonals \( AD \), \( BE \), and \( CF \) of a hexagon \( ABCDEF \) inscribed in a circle \( k \) intersect at a point \( P \), and the acute angle between any two of them is \( 60^\circ \). Let \( r_{AB} \) be the radius of the circle tangent to segments \( PA \) and \( PB \) and internally tangent to \( k \); the radii \( r_{BC} \), \( r_{CD} \), \( r_{DE} \), \( r_{EF} \), and \( r_{FA} \) are defined similarly. Prove that
\[
r_{AB}r_{CD} + r_{CD}r_{EF} + r_{EF}r_{AB} = r_{BC}r_{DE} + r_{DE}r_{FA} + r_{FA}r_{BC}.
\]
2 replies
GeorgeRP
Sep 10, 2024
Assassino9931
Yesterday at 7:00 PM
Interesting functional equation with geometry
User21837561   3
N Yesterday at 6:34 PM by Double07
Source: BMOSL 2024 G7
For an acute triangle $ABC$, let $O$ be the circumcentre, $H$ be the orthocentre, and $G$ be the centroid.
Let $f:\pi\rightarrow\mathbb R$ satisfy the following condition:
$f(A)+f(B)+f(C)=f(O)+f(G)+f(H)$
Prove that $f$ is constant.
3 replies
User21837561
Yesterday at 8:14 AM
Double07
Yesterday at 6:34 PM
greatest volume
hzbrl   1
N Yesterday at 6:32 PM by hzbrl
Source: purple comet
A large sphere with radius 7 contains three smaller balls each with radius 3 . The three balls are each externally tangent to the other two balls and internally tangent to the large sphere. There are four right circular cones that can be inscribed in the large sphere in such a way that the bases of the cones are tangent to all three balls. Of these four cones, the one with the greatest volume has volume $n \pi$. Find $n$.
1 reply
hzbrl
Thursday at 9:56 AM
hzbrl
Yesterday at 6:32 PM
(n+1)2^n, (n+3)2^{n+2} not perfect squares for the same n
parmenides51   3
N Yesterday at 6:31 PM by AylyGayypow009
Source: Greece JBMO TST 2015 p3
Prove that there is not a positive integer $n$ such that numbers $(n+1)2^n, (n+3)2^{n+2}$ are both perfect squares.
3 replies
parmenides51
Apr 29, 2019
AylyGayypow009
Yesterday at 6:31 PM
Some nice summations
amitwa.exe   31
N Apr 23, 2025 by soryn
Problem 1: $\Omega=\left(\sum_{0\le i\le j\le k}^{\infty} \frac{1}{3^i\cdot4^j\cdot5^k}\right)\left(\mathop{{\sum_{i=0}^{\infty}\sum_{j=0}^{\infty}\sum_{k=0}^{\infty}}}_{i\neq j\neq k}\frac{1}{3^i\cdot3^j\cdot3^k}\right)=?$
31 replies
amitwa.exe
May 24, 2024
soryn
Apr 23, 2025
Some nice summations
G H J
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amitwa.exe
347 posts
#1 • 2 Y
Y by P162008, cubres
Problem 1: $\Omega=\left(\sum_{0\le i\le j\le k}^{\infty} \frac{1}{3^i\cdot4^j\cdot5^k}\right)\left(\mathop{{\sum_{i=0}^{\infty}\sum_{j=0}^{\infty}\sum_{k=0}^{\infty}}}_{i\neq j\neq k}\frac{1}{3^i\cdot3^j\cdot3^k}\right)=?$
This post has been edited 4 times. Last edited by amitwa.exe, Aug 6, 2024, 5:43 AM
Z K Y
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amitwa.exe
347 posts
#2 • 2 Y
Y by P162008, cubres
Problem 2: Given
$$\Omega_1=\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}\sum_{k=1}^{\infty}\frac{ijk(3i^2j+ijk)\cdot5^{-(i+j+k)}}{(i+j)(j+k)(i+k)},$$$$\Omega_2=\sum_{i,j\ge 0,i+j=\text{odd}}\frac{1}{3^i\cdot5^j}, \Omega_3=\sum_{i,j\ge 0,i+j=\text{even}}\frac{1}{3^i\cdot5^j}$$Find ${\Omega_1}^{\Omega_2+\Omega_3}$
This post has been edited 2 times. Last edited by amitwa.exe, May 25, 2024, 3:21 AM
Z K Y
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amitwa.exe
347 posts
#3 • 1 Y
Y by cubres
Problem 3: Let each $x_i (i=1,2,3,\cdots,2020)$ be a positive real number such that $\sum_{i=1}^{2020} x_i=1$ and $\sum_{n=1}^{2020}\sum_{m=1}^{\infty} x_n^m=k$. Then find the value of the following summation in terms of k.
$$\sum_{n=1}^{2020}\sum_{m=2}^{\infty} x_n^m$$
This post has been edited 2 times. Last edited by amitwa.exe, Dec 9, 2024, 6:35 AM
Z K Y
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amitwa.exe
347 posts
#4 • 1 Y
Y by cubres
Problem 4: Evaluate $S=\sum_{k=1}^{2018} \left\lfloor{\frac{103k}{2019}}\right\rfloor$ (where $\left\lfloor a \right\rfloor$ is the greatest integer less then or equal $a$)
This post has been edited 1 time. Last edited by amitwa.exe, May 24, 2024, 8:56 AM
Z K Y
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amitwa.exe
347 posts
#5 • 1 Y
Y by cubres
Problem 5: Let $\{a_n\}$ be a sequence defined by $a_1=a_2=2$ and $\frac{a_{n-1}a_n}{a_{n-1}a_{n+1}-a_n^2}=\frac{8n^3+12n^2-2n-3}{4}, \forall n\ge 2$. Find the least integer $m$ satisfying $\sum_{n=2}^{2020}\frac{a_{n+1}}{a_n}\le m$
This post has been edited 1 time. Last edited by amitwa.exe, May 24, 2024, 8:56 AM
Z K Y
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amitwa.exe
347 posts
#6 • 1 Y
Y by cubres
Problem 6: Prove that $\sum_{r=0}^{\left\lfloor\frac{n}{2}\right\rfloor}\binom{n}{2r+1}5^r=2^{n-1}F_n$, where $F_n$ is n-th Fibonacci number.
This post has been edited 1 time. Last edited by amitwa.exe, May 24, 2024, 8:56 AM
Z K Y
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amitwa.exe
347 posts
#7 • 1 Y
Y by cubres
Problem 7: A sequence ${a_n}$ of real numbers is defined by: $a_1=1, a_2=2020$ and $\frac{a_n}{a_{n-1}}=\frac{(n+1)a_{n-1}}{a_{n-1}+(n+1)a_{n-2}} \forall n\ge 3$. Find the least natural number $k$ such that $\left\lfloor{\frac{a_k}{a_{k-1}}}\right\rfloor=0.$ (where $\left\lfloor a \right\rfloor$ is the greatest integer less then or equal $a$).
This post has been edited 1 time. Last edited by amitwa.exe, May 24, 2024, 8:55 AM
Z K Y
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amitwa.exe
347 posts
#8 • 2 Y
Y by cubres, P162008
Problem 8: Evaluate $\sum_{n=2}^{\infty}\sum_{k=2}^{\infty} \frac{1}{k^n\cdot k!}$
This post has been edited 1 time. Last edited by amitwa.exe, May 25, 2024, 4:46 AM
Reason: corrected the lower limit
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aaravdodhia
2602 posts
#9 • 1 Y
Y by cubres
Problem 1: In the second part $1/3^{i+j+k}$ appears $\binom{i+j+k+2}{2}$ times, so it's $$9\sum_{s=2}^{\infty} \binom{s}{2}/3^S = 9/2 \sum [s^2/3^s - s/3^s] = 9/2 \cdot (3/2 - 3/4) = 27/4.$$
We can break the first part into independent series: $$\sum_{i=0}^\infty 1/3^i \sum_{j=i}^\infty 1/4^j \sum_{k=j}^\infty 1/5^k = \frac{5}{4} \sum 1/3^i \sum_{j=i}^\infty 1/5^j = \left(\frac54\right)^2 \sum_{i=0}^\infty 1/3^i\cdot 1/5^i = 375/224.$$So the answer is $10,125/896$?
This post has been edited 1 time. Last edited by aaravdodhia, May 24, 2024, 10:04 PM
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phungthienphuoc
95 posts
#10 • 2 Y
Y by cubres, soryn
Problem 1: $\Omega = \dfrac{30\,375}{58\,292}$
Solution of problem 1
This post has been edited 2 times. Last edited by phungthienphuoc, May 25, 2024, 2:40 AM
Z K Y
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amitwa.exe
347 posts
#11 • 2 Y
Y by phungthienphuoc, cubres
phungthienphuoc wrote:
Problem 1: $\Omega = \dfrac{30\,375}{58\,292}$
Solution of problem 1

Correct
Z K Y
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RedFireTruck
4223 posts
#12 • 3 Y
Y by amitwa.exe, cubres, soryn
amitwa.exe wrote:
Problem 4: Evaluate $S=\sum_{k=1}^{2018} \left\lfloor{\frac{103k}{2019}}\right\rfloor$ (where $\left\lfloor a \right\rfloor$ is the greatest integer less then or equal $a$)

$S=\sum_{k=1}^{2018} \left\lfloor{\frac{103k}{2019}}\right\rfloor=\sum_{k=1}^{2018} \frac{103k}{2019}-\sum_{k=1}^{2018} \{\frac{103k}{2019}\}=\frac{103}{2019}\frac{2018\cdot 2019}{2}-\sum_{k=1}^{2018} \frac{k}{2019}=103\cdot 1009-\frac{1}{2019}\frac{2018\cdot2019}{2}=102\cdot1009=\boxed{102918}$
Z K Y
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phungthienphuoc
95 posts
#13 • 1 Y
Y by cubres
amitwa.exe wrote:
Problem 2: Given
$$\Omega_1=\sum_{i=0}^{\infty}\sum_{j=0}^{\infty}\sum_{k=0}^{\infty}\frac{ijk(3i^2j+ijk)\cdot5^{-(i+j+k)}}{(i+j)(j+k)(i+k)},$$$$\Omega_2=\sum_{i,j\ge 0,i+j=\text{odd}}\frac{1}{3^i\cdot5^j}, \Omega_3=\sum_{i,j\ge 0,i+j=\text{even}}\frac{1}{3^i\cdot5^j}$$Find ${\Omega_1}^{\Omega_2+\Omega_3}$

In case $i=j=0$, $\Omega_1$ is undefined.

By the way, I think you mistyped problem 3.
This post has been edited 1 time. Last edited by phungthienphuoc, May 25, 2024, 3:20 AM
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amitwa.exe
347 posts
#14 • 2 Y
Y by phungthienphuoc, cubres
phungthienphuoc wrote:
amitwa.exe wrote:
Problem 2: Given
$$\Omega_1=\sum_{i=0}^{\infty}\sum_{j=0}^{\infty}\sum_{k=0}^{\infty}\frac{ijk(3i^2j+ijk)\cdot5^{-(i+j+k)}}{(i+j)(j+k)(i+k)},$$$$\Omega_2=\sum_{i,j\ge 0,i+j=\text{odd}}\frac{1}{3^i\cdot5^j}, \Omega_3=\sum_{i,j\ge 0,i+j=\text{even}}\frac{1}{3^i\cdot5^j}$$Find ${\Omega_1}^{\Omega_2+\Omega_3}$

In case $i=j=0$, $\Omega_1$ is undefined.

Thanks for correction
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ABCDTNT__
144 posts
#15 • 2 Y
Y by cubres, soryn
amitwa.exe wrote:
Problem 8: Evaluate $\sum_{n=2}^{\infty}\sum_{n=2}^{\infty} \frac{1}{k^n\cdot k!}$

Did you mistype the $k=2$ under the $\Sigma$ into $n=2$? Well, i think i can give a solution after correction :maybe:

$$\begin{aligned}
\sum_{k=2}^{\infty}\sum_{n=2}^{\infty}\frac{1}{k^n\cdot k!} &= \sum_{k=2}^{\infty}\frac{1}{k!}\cdot\left (\sum_{n=2}^{\infty}\left (\frac{1}{k}\right )^n\right ) =\sum_{k=2}^{\infty}\frac{1}{k!}\cdot\frac{1}{k(k-1)} = \sum_{k=2}^{\infty}\frac{1}{k!}\left ( \frac{1}{k-1} - \frac{1}{k}\right )\\
&= \sum_{k=2}^{\infty}\left ( \frac{1}{k!\cdot(k-1)}-\frac{1}{k!\cdot k}\right )=\sum_{k=2}^{\infty}\frac{1}{k!\cdot(k-1)}-\sum_{k=2}^{\infty}\frac{1}{k!\cdot k}\\
&= \sum_{k=1}^{\infty}\frac{1}{(k+1)!\cdot k}-\sum_{k=2}^{\infty}\frac{1}{k!\cdot k} = \frac{1}{1\cdot 2!}+\sum_{k=2}^{\infty}\left ( \frac{1-(k+1)}{(k+1)!\cdot k}\right )\\
&= \frac{1}{2}-\sum_{k=2}^{\infty}\frac{1}{(k+1)!} = \frac{1}{2} - (e-1-1-\frac{1}{2}) = 3-e
\end{aligned}
$$In the last step we used $\sum_{k=0}^{\infty}\frac{1}{k!}=e$ (according to Taylor expansion). So the answer is $\boxed{3-e}$. $\square$
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amitwa.exe
347 posts
#16 • 1 Y
Y by cubres
nicely done
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ABCDTNT__
144 posts
#17 • 3 Y
Y by amitwa.exe, cubres, soryn
amitwa.exe wrote:
Problem 7: A sequence ${a_n}$ of real numbers is defined by: $a_1=1, a_2=2020$ and $\frac{a_n}{a_{n-1}}=\frac{(n+1)a_{n-1}}{a_{n-1}+(n+1)a_{n-2}} \forall n\ge 3$. Find the least natural number $k$ such that $\left\lfloor{\frac{a_k}{a_{k-1}}}\right\rfloor=0.$ (where $\left\lfloor a \right\rfloor$ is the greatest integer less then or equal $a$).

Take the reciprocal of both sides of the original formula, we get
$$ \frac{a_{n-1}}{a_n} = \frac{a_{n-1}+(n+1)a_{n-2}}{(n+1)a_{n-1}}=\frac{1}{n+1}+\frac{a_{n-2}}{a_{n-1}} $$Summarize it, we have
$$\frac{a_{n-1}}{a_n} = \frac{1}{n+1}+\frac{1}{n}+\cdots+\frac{1}{4}+\frac{a_1}{a_2}=\frac{1}{2020}+\left (\frac{1}{4}+\frac{1}{5}+\cdots+\frac{1}{n+1}\right )$$On the other hand,
$$\left\lfloor \frac{a_n}{a_{n-1}} \right\rfloor=0 \Leftrightarrow 0\le \frac{a_n}{a_{n-1}} < 1 \Leftrightarrow \frac{a_{n-1}}{a_n}>1$$and through calculation we get
$$\frac{a_7}{a_8}=\frac{50707}{50904}<1 , \frac{a_8}{a_9}=\frac{278987}{254520}>1$$so the answer is $\boxed{k=9}$ (since the sequence $\left \{a_n\right \}$ is obviously increasing). $\square$
This post has been edited 1 time. Last edited by ABCDTNT__, May 25, 2024, 6:16 AM
Reason: Grammar mistakes.
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melowmolly
502 posts
#18 • 2 Y
Y by cubres, soryn
amitwa.exe wrote:
Problem 5: Let $\{a_n\}$ be a sequence defined by $a_1=a_2=2$ and $\frac{a_{n-1}a_n}{a_{n-1}a_{n+1}-a_n^2}=\frac{8n^3+12n^2-2n-3}{4}, \forall n\ge 2$. Find the least integer $m$ satisfying $\sum_{n=2}^{2020}\frac{a_{n+1}}{a_n}\le m$

Note that \[\frac{a_{n-1}a_{n+1}-a_n^2}{a_{n-1}a_n}=\frac{a_{n+1}}{a_n}-\frac{a_n}{a_{n-1}}=\frac{4}{8n^3+12n^2-2n-3}=\frac{4}{(2n+3)(2n+1)(2n-1)}=\frac{1}{(2n+1)(2n-1)}-\frac{1}{(2n+3)(2n+1)}\]\[\Rightarrow \frac{a_{n+1}}{a_n}+\frac{1}{(2n+3)(2n+1)}=\frac{a_n}{a_{n-1}}+\frac{1}{(2n+1)(2n-1)}=\dots=\frac{a_2}{a_1}+\frac{1}{5\cdot 3}=\frac{16}{15}\]So \[\sum_{n=2}^{2020}\frac{a_{n+1}}{a_n}=\sum_{n=2}^{2020}\left(\frac{16}{15}-\frac1{(2n+3)(2n+1)}\right)=\frac{10768}5-\frac12\sum_{n=2}^{2020}\left(\frac1{2n+1}-\frac1{2n+3}\right)=\frac{10768}5-\frac{1}2\left(\frac{1}5-\frac1{4043}\right).\]It follows immediately that \[m=2154.\]
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aman_maths
34 posts
#19 • 1 Y
Y by cubres
S3
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phungthienphuoc
95 posts
#20 • 1 Y
Y by cubres
amitwa.exe wrote:
Problem 6: Prove that $\sum_{r=0}^{\left\lfloor\frac{n}{2}\right\rfloor}\binom{n}{2r+1}5^r=2^{n-1}F_n$, where $F_n$ is n-th Fibonacci number.

$$\begin{matrix}
\displaystyle
\sum_{r=0}^{\left\lfloor\frac{n}{2}\right\rfloor}\binom{n}{2r+1}5^r
&=&
\displaystyle
\dfrac{1}{\sqrt{5}}\sum_{r=0}^{\left\lfloor\frac{n}{2}\right\rfloor}\binom{n}{2r+1}(\sqrt{5})^{2r+1}
\\\\
&=&
\dfrac{1}{\sqrt{5}}\dfrac{(1+\sqrt{5})^n-(1-\sqrt{5})^n}{2}
\\\\
&=&
2^{n-1}\dfrac{\left(\dfrac{1+\sqrt{5}}{2}\right)^n-\left(\dfrac{1-\sqrt{5}}{2}\right)^n}{\sqrt{5}}
\\\\
&=&
2^{n-1}F_n
\end{matrix}$$
This post has been edited 1 time. Last edited by phungthienphuoc, May 25, 2024, 8:16 PM
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amitwa.exe
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#21 • 1 Y
Y by cubres
Problem 9: Let $f(x)$ be a polynomial of 6 degree satisfying $f(2021)=2021$ and $f(1)=f(3)=f(5)=f(2016)=f(2018)=f(2020)=2020$
Then find $\Omega=f(1)-f(2)+f(3)-f(4)+\cdots+f(2019)-f(2020)+f(2021)$
This post has been edited 1 time. Last edited by amitwa.exe, May 26, 2024, 10:26 AM
Reason: added -f(2020)+f(2021) at the end
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amitwa.exe
347 posts
#22 • 1 Y
Y by cubres
Problem 10: Evaluate : $\frac{\displaystyle\sum_{k=1}^{n^2-1}\sqrt{n+\sqrt{k}}}{\displaystyle\sum_{k=1}^{n^2-1}\sqrt{n-\sqrt{k}}}$
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amitwa.exe
347 posts
#23 • 1 Y
Y by cubres
Problem 11: A sequence $\{a\}$ is defined by $a_0=1=a_1$ and $a_na_{n-2}=a_{n-1}a_{n-2}+a_{n-1}^2$ for $n\ge 2$. If $a_n\neq 0, \forall n$, then find the value of
$$\sum_{n=1}^{2020} na_n$$
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ABCDTNT__
144 posts
#24 • 2 Y
Y by cubres, soryn
amitwa.exe wrote:
Problem 9: Let $f(x)$ be a polynomial of 6 degree satisfying $f(2021)=2021$ and $f(1)=f(3)=f(5)=f(2016)=f(2018)=f(2020)=2020$
Then find $\Omega=f(1)-f(2)+f(3)-f(4)+\cdots+f(2019)$

Since $deg f=6$, suppose $f(x)=a(x-1)(x-3)(x-5)(x-2016)(x-2018)(x-2020)+2020$ (where $a\in \mathbb{R}/{0}$)
Then we have $f(2021-x)=a(2020-x)(2018-x)(2016-x)(5-x)(3-x)(1-x)+2020=f(x)$, which means $f(x)-f(2021-x)=0$.
Hence, $\Omega=f(1)+(-f(2)+f(2019))+(f(3)-f(2018))+\cdots+(-f(1010)+f(1011))=f(1)=2020$.

Is there anything wrong? I noticed that i didn't use the $f(2021)=2021$ in the problem.
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amitwa.exe
347 posts
#25 • 1 Y
Y by cubres
ABCDTNT__ wrote:
amitwa.exe wrote:
Problem 9: Let $f(x)$ be a polynomial of 6 degree satisfying $f(2021)=2021$ and $f(1)=f(3)=f(5)=f(2016)=f(2018)=f(2020)=2020$
Then find $\Omega=f(1)-f(2)+f(3)-f(4)+\cdots+f(2019)$

Since $deg f=6$, suppose $f(x)=a(x-1)(x-3)(x-5)(x-2016)(x-2018)(x-2020)+2020$ (where $a\in \mathbb{R}/{0}$)
Then we have $f(2021-x)=a(2020-x)(2018-x)(2016-x)(5-x)(3-x)(1-x)+2020=f(x)$, which means $f(x)-f(2021-x)=0$.
Hence, $\Omega=f(1)+(-f(2)+f(2019))+(f(3)-f(2018))+\cdots+(-f(1010)+f(1011))=f(1)=2020$.

Is there anything wrong? I noticed that i didn't use the $f(2021)=2021$ in the problem.

i actually forgot to type $-f(2020)+f(2021)$ at the end :blush:
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ABCDTNT__
144 posts
#26 • 2 Y
Y by cubres, soryn
amitwa.exe wrote:
i actually forgot to type $-f(2020)+f(2021)$ at the end :blush:

Fortunately, the problem is still able to be solved :-D only small matters, don't care too much.
amitwa.exe wrote:
Problem 11: A sequence $\{a\}$ is defined by $a_0=1=a_1$ and $a_na_{n-2}=a_{n-1}a_{n-2}+a_{n-1}^2$ for $n\ge 2$. If $a_n\neq 0, \forall n$, then find the value of
$$\sum_{n=1}^{2020} na_n$$

Since $a_n\neq 0,\forall n\ge 2$, divide both side of the equation by $a_{n-1}a_{n-2}$, we have
$$\frac{a_n}{a_{n-1}}=1+\frac{a_{n-1}}{a_{n-2}}$$which leads to
$$\frac{a_n}{a_{n-1}}=n-1+\frac{a_1}{a_0}=n$$hence
$$a_n=n!, \forall n\ge 1$$so that we can infer that
$$\sum_{n=1}^{2020}na_n=\sum_{n=1}^{2020}n\cdot n!=\sum_{n=1}^{2020}\left ((n+1)!-n!\right )=2021!-1$$In summary, the answer is $\boxed{2021!-1}$. $\square$
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ABCDTNT__
144 posts
#27 • 2 Y
Y by cubres, soryn
amitwa.exe wrote:
Problem 2: Given
$$\Omega_1=\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}\sum_{k=1}^{\infty}\frac{ijk(3i^2j+ijk)\cdot5^{-(i+j+k)}}{(i+j)(j+k)(i+k)},$$$$\Omega_2=\sum_{i,j\ge 0,i+j=\text{odd}}\frac{1}{3^i\cdot5^j}, \Omega_3=\sum_{i,j\ge 0,i+j=\text{even}}\frac{1}{3^i\cdot5^j}$$Find ${\Omega_1}^{\Omega_2+\Omega_3}$

:( The answer is not that pretty as i hoped. Maybe i made some mistakes :(

Notice that the $i, j,k$ in $\Omega_1$ has the same position. It implies
$$\begin{aligned}
\Omega_1&=\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}\sum_{k=1}^{\infty}\frac{ijk(3i^2j+ijk)\cdot5^{-(i+j+k)}}{(i+j)(j+k)(i+k)}
=\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}\sum_{k=1}^{\infty}\frac{ijk(3j^2i+ijk)\cdot5^{-(i+j+k)}}{(i+j)(j+k)(i+k)}\\
&=\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}\sum_{k=1}^{\infty}\frac{ijk(3j^2k+ijk)\cdot5^{-(i+j+k)}}{(i+j)(j+k)(i+k)}
=\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}\sum_{k=1}^{\infty}\frac{ijk(3k^2j+ijk)\cdot5^{-(i+j+k)}}{(i+j)(j+k)(i+k)}\\
&=\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}\sum_{k=1}^{\infty}\frac{ijk(3k^2i+ijk)\cdot5^{-(i+j+k)}}{(i+j)(j+k)(i+k)}
=\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}\sum_{k=1}^{\infty}\frac{ijk(3i^2k+ijk)\cdot5^{-(i+j+k)}}{(i+j)(j+k)(i+k)}
\end{aligned}$$Add them up, then it comes to
$$\begin{aligned}
\Omega_1&=\frac{1}{2}\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}\sum_{k=1}^{\infty}\frac{ijk(i^2j+ij^2+j^2k+jk^2+k^2i+ki^2+2ijk)\cdot5^{-(i+j+k)}}{(i+j)(j+k)(i+k)}\\
&=\frac{1}{2}\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}\sum_{k=1}^{\infty}\frac{ijk(i+j)(j+k)(k+i)\cdot5^{-(i+j+k)}}{(i+j)(j+k)(i+k)}\\
&=\frac{1}{2}\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}\sum_{k=1}^{\infty}ijk\cdot5^{-(i+j+k)}
\end{aligned}$$Let $A=\sum_{k=1}^{\infty}\frac{k}{5^k}$, we can easily get $A=\frac{5}{16}$. In that case,
$$\Omega_1=\frac{1}{2}\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}\frac{ij}{5^{i+j}}\cdot\sum_{k=1}^{\infty}\frac{k}{5^k}=\frac{1}{2}\cdot\frac{5}{16}\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}\frac{ij}{5^{i+j}}=\cdots=\frac{1}{2}\cdot\left ( \frac{5}{16}\right ) ^3$$
On the other hand,
$$\begin{aligned}
\Omega_2+\Omega_3&=\sum_{i,j\ge 0}\frac{1}{3^i\cdot5^j}=\sum_{i=0}^{\infty}\sum_{j=0}^{\infty}\frac{1}{3^i\cdot5^j}\\
&=\left (\sum_{i=0}^{\infty}\frac{1}{3^i}\right )\left (\sum_{j=0}^{\infty}\frac{1}{5^j}\right )=\frac{3}{2}\cdot\frac{5}{4}=\frac{15}{8}
\end{aligned}$$Hence the answer $\Omega_1^{\Omega_2+\Omega_3}=\left (\frac{1}{2}\cdot\left (\frac{5}{16}\right )^3\right )^\frac{15}{8}=\boxed{\frac{5^\frac{45}{8}}{2^\frac{195}{8}}}$. $\square$
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amitwa.exe
347 posts
#28 • 1 Y
Y by cubres
It is actually correct :10: . Nice solution
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Aiden-1089
285 posts
#29 • 2 Y
Y by cubres, soryn
amitwa.exe wrote:
Problem 10: Evaluate : $\frac{\displaystyle\sum_{k=1}^{n^2-1}\sqrt{n+\sqrt{k}}}{\displaystyle\sum_{k=1}^{n^2-1}\sqrt{n-\sqrt{k}}}$
First we see that $\left( \sqrt{n+\sqrt{k}} - \sqrt{n-\sqrt{k}} \right) ^2 = 2n - 2\sqrt{n^2-k}$.
Now,
$\sum_{k=1}^{n^2-1}\sqrt{n+\sqrt{k}} - \sum_{k=1}^{n^2-1}\sqrt{n-\sqrt{k}} = \sum_{k=1}^{n^2-1} \left( \sqrt{n+\sqrt{k}} - \sqrt{n-\sqrt{k}} \right)$
$= \sum_{k=1}^{n^2-1} \sqrt{2n - 2\sqrt{n^2-k}} = \sqrt{2} \sum_{k=1}^{n^2-1} \sqrt{n - \sqrt{n^2-k}} = \sqrt{2} \sum_{k=1}^{n^2-1} \sqrt{n - \sqrt{k}}$.
So $\frac{\displaystyle\sum_{k=1}^{n^2-1}\sqrt{n+\sqrt{k}}}{\displaystyle\sum_{k=1}^{n^2-1}\sqrt{n-\sqrt{k}}}=\boxed{1+\sqrt{2}}$.
amitwa.exe wrote:
Problem 11: A sequence $\{a\}$ is defined by $a_0=1=a_1$ and $a_na_{n-2}=a_{n-1}a_{n-2}+a_{n-1}^2$ for $n\ge 2$. If $a_n\neq 0, \forall n$, then find the value of
$$\sum_{n=1}^{2020} na_n$$
$a_na_{n-2}=a_{n-1}a_{n-2}+a_{n-1}^2 \implies a_n=a_{n-1}+\frac{a_{n-1}^2}{a_{n-2}} \implies \frac{a_n}{a_{n-1}}=1+\frac{a_{n-1}}{a_{n-2}}$. Since $\frac{a_1}{a_0}=1$, we get that $\frac{a_n}{a_{n-1}}=n$ for all $n \geq 1$. Now $a_0=1$, so $a_n=n!$ for all $n \geq 0$.
$\sum_{n=1}^{k} na_n = \sum_{n=1}^{k} n \cdot n! = \sum_{n=1}^{k} \left( (n+1)! - n! \right) = (k+1)! -1$, so $\sum_{n=1}^{2020} na_n = \boxed{2021!-1}$.
This post has been edited 1 time. Last edited by Aiden-1089, May 26, 2024, 5:20 PM
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P162008
187 posts
#31 • 2 Y
Y by teomihai, soryn
amitwa.exe wrote:
Problem 2: Given
$$\Omega_1=\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}\sum_{k=1}^{\infty}\frac{ijk(3i^2j+ijk)\cdot5^{-(i+j+k)}}{(i+j)(j+k)(i+k)},$$$$\Omega_2=\sum_{i,j\ge 0,i+j=\text{odd}}\frac{1}{3^i\cdot5^j}, \Omega_3=\sum_{i,j\ge 0,i+j=\text{even}}\frac{1}{3^i\cdot5^j}$$Find ${\Omega_1}^{\Omega_2+\Omega_3}$

Solution
This post has been edited 7 times. Last edited by P162008, Apr 29, 2025, 8:58 PM
Reason: Typo
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P162008
187 posts
#32 • 1 Y
Y by soryn
amitwa.exe wrote:
Problem 8: Evaluate $\sum_{n=2}^{\infty}\sum_{k=2}^{\infty} \frac{1}{k^n\cdot k!}$

Solution
This post has been edited 9 times. Last edited by P162008, Apr 28, 2025, 9:35 PM
Reason: Typo
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soryn
5342 posts
#33
Y by
Very,very nice! Good job! Congratulations for all.. Instructivelu for me...
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