Stay ahead of learning milestones! Enroll in a class over the summer!

G
Topic
First Poster
Last Poster
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)!

Be sure to mark your calendars for the following upcoming events:
[list][*]May 9th, 4:30pm PT/7:30pm ET, Casework 2: Overwhelming Evidence — A Text Adventure, a game where participants will work together to navigate the map, solve puzzles, and win! All are welcome.
[*]May 19th, 4:30pm PT/7:30pm ET, What's Next After Beast Academy?, designed for students finishing Beast Academy and ready for Prealgebra 1.
[*]May 20th, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 1 Math Jam, Problems 1 to 4, join the Canada/USA Mathcamp staff for this exciting Math Jam, where they discuss solutions to Problems 1 to 4 of the 2025 Mathcamp Qualifying Quiz!
[*]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]
Our full course list for upcoming classes is below:
All classes run 7:30pm-8:45pm ET/4:30pm - 5:45pm PT unless otherwise noted.

Introductory: Grades 5-10

Prealgebra 1 Self-Paced

Prealgebra 1
Tuesday, May 13 - Aug 26
Thursday, May 29 - Sep 11
Sunday, Jun 15 - Oct 12
Monday, Jun 30 - Oct 20
Wednesday, Jul 16 - Oct 29

Prealgebra 2 Self-Paced

Prealgebra 2
Wednesday, May 7 - Aug 20
Monday, Jun 2 - Sep 22
Sunday, Jun 29 - Oct 26
Friday, Jul 25 - Nov 21

Introduction to Algebra A Self-Paced

Introduction to Algebra A
Sunday, May 11 - Sep 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Wednesday, May 14 - Aug 27
Friday, May 30 - Sep 26
Monday, Jun 2 - Sep 22
Sunday, Jun 15 - Oct 12
Thursday, Jun 26 - Oct 9
Tuesday, Jul 15 - Oct 28

Introduction to Counting & Probability Self-Paced

Introduction to Counting & Probability
Thursday, May 15 - Jul 31
Sunday, Jun 1 - Aug 24
Thursday, Jun 12 - Aug 28
Wednesday, Jul 9 - Sep 24
Sunday, Jul 27 - Oct 19

Introduction to Number Theory
Friday, May 9 - Aug 1
Wednesday, May 21 - Aug 6
Monday, Jun 9 - Aug 25
Sunday, Jun 15 - Sep 14
Tuesday, Jul 15 - Sep 30

Introduction to Algebra B Self-Paced

Introduction to Algebra B
Tuesday, May 6 - Aug 19
Wednesday, Jun 4 - Sep 17
Sunday, Jun 22 - Oct 19
Friday, Jul 18 - Nov 14

Introduction to Geometry
Sunday, May 11 - Nov 9
Tuesday, May 20 - Oct 28
Monday, Jun 16 - Dec 8
Friday, Jun 20 - Jan 9
Sunday, Jun 29 - Jan 11
Monday, Jul 14 - Jan 19

Paradoxes and Infinity
Mon, Tue, Wed, & Thurs, Jul 14 - Jul 16 (meets every day of the week!)

Intermediate: Grades 8-12

Intermediate Algebra
Sunday, Jun 1 - Nov 23
Tuesday, Jun 10 - Nov 18
Wednesday, Jun 25 - Dec 10
Sunday, Jul 13 - Jan 18
Thursday, Jul 24 - Jan 22

Intermediate Counting & Probability
Wednesday, May 21 - Sep 17
Sunday, Jun 22 - Nov 2

Intermediate Number Theory
Sunday, Jun 1 - Aug 24
Wednesday, Jun 18 - Sep 3

Precalculus
Friday, May 16 - Oct 24
Sunday, Jun 1 - Nov 9
Monday, Jun 30 - Dec 8

Advanced: Grades 9-12

Olympiad Geometry
Tuesday, Jun 10 - Aug 26

Calculus
Tuesday, May 27 - Nov 11
Wednesday, Jun 25 - Dec 17

Group Theory
Thursday, Jun 12 - Sep 11

Contest Preparation: Grades 6-12

MATHCOUNTS/AMC 8 Basics
Friday, May 23 - Aug 15
Monday, Jun 2 - Aug 18
Thursday, Jun 12 - Aug 28
Sunday, Jun 22 - Sep 21
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

MATHCOUNTS/AMC 8 Advanced
Sunday, May 11 - Aug 10
Tuesday, May 27 - Aug 12
Wednesday, Jun 11 - Aug 27
Sunday, Jun 22 - Sep 21
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

AMC 10 Problem Series
Friday, May 9 - Aug 1
Sunday, Jun 1 - Aug 24
Thursday, Jun 12 - Aug 28
Tuesday, Jun 17 - Sep 2
Sunday, Jun 22 - Sep 21 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Monday, Jun 23 - Sep 15
Tues & Thurs, Jul 8 - Aug 14 (meets twice a week!)

AMC 10 Final Fives
Sunday, May 11 - Jun 8
Tuesday, May 27 - Jun 17
Monday, Jun 30 - Jul 21

AMC 12 Problem Series
Tuesday, May 27 - Aug 12
Thursday, Jun 12 - Aug 28
Sunday, Jun 22 - Sep 21
Wednesday, Aug 6 - Oct 22

AMC 12 Final Fives
Sunday, May 18 - Jun 15

AIME Problem Series A
Thursday, May 22 - Jul 31

AIME Problem Series B
Sunday, Jun 22 - Sep 21

F=ma Problem Series
Wednesday, Jun 11 - Aug 27

WOOT Programs
Visit the pages linked for full schedule details for each of these programs!


MathWOOT Level 1
MathWOOT Level 2
ChemWOOT
CodeWOOT
PhysicsWOOT

Programming

Introduction to Programming with Python
Thursday, May 22 - Aug 7
Sunday, Jun 15 - Sep 14 (1:00 - 2:30 pm ET/10:00 - 11:30 am PT)
Tuesday, Jun 17 - Sep 2
Monday, Jun 30 - Sep 22

Intermediate Programming with Python
Sunday, Jun 1 - Aug 24
Monday, Jun 30 - Sep 22

USACO Bronze Problem Series
Tuesday, May 13 - Jul 29
Sunday, Jun 22 - Sep 1

Physics

Introduction to Physics
Wednesday, May 21 - Aug 6
Sunday, Jun 15 - Sep 14
Monday, Jun 23 - Sep 15

Physics 1: Mechanics
Thursday, May 22 - Oct 30
Monday, Jun 23 - Dec 15

Relativity
Mon, Tue, Wed & Thurs, Jun 23 - Jun 26 (meets every day of the week!)
0 replies
jlacosta
May 1, 2025
0 replies
Only n-digits 1234
Assassino9931   10
N a few seconds ago by MELSSATIMOV40
Source: Balkan MO Shortlist 2024 N1
Let $n$ be a positive integer. Do there exist distinct $n$-digit positive integers $x$ and $y$, with each of their digits being $1,2,3$ or $4$, such that $4^n$ divides $x-y$?
10 replies
Assassino9931
Apr 27, 2025
MELSSATIMOV40
a few seconds ago
hard problem
Cobedangiu   4
N 14 minutes ago by Cobedangiu
$a,b,c>0$ and $a+b+c=7$. CM:
$\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}+abc \ge ab+bc+ca-2$
4 replies
+1 w
Cobedangiu
Yesterday at 4:24 PM
Cobedangiu
14 minutes ago
Aime type Geo
ehuseyinyigit   1
N 23 minutes ago by ehuseyinyigit
Source: Turkish First Round 2024
In a scalene triangle $ABC$, let $M$ be the midpoint of side $BC$. Let the line perpendicular to $AC$ at point $C$ intersect $AM$ at $N$. If $(BMN)$ is tangent to $AB$ at $B$, find $AB/MA$.
1 reply
ehuseyinyigit
Yesterday at 9:04 PM
ehuseyinyigit
23 minutes ago
Arbitrary point on BC and its relation with orthocenter
falantrng   34
N 26 minutes ago by Mamadi
Source: Balkan MO 2025 P2
In an acute-angled triangle \(ABC\), \(H\) be the orthocenter of it and \(D\) be any point on the side \(BC\). The points \(E, F\) are on the segments \(AB, AC\), respectively, such that the points \(A, B, D, F\) and \(A, C, D, E\) are cyclic. The segments \(BF\) and \(CE\) intersect at \(P.\) \(L\) is a point on \(HA\) such that \(LC\) is tangent to the circumcircle of triangle \(PBC\) at \(C.\) \(BH\) and \(CP\) intersect at \(X\). Prove that the points \(D, X, \) and \(L\) lie on the same line.

Proposed by Theoklitos Parayiou, Cyprus
34 replies
falantrng
Apr 27, 2025
Mamadi
26 minutes ago
Inequalities
sqing   8
N 5 hours ago by sqing
Let $ a,b,c>0 $ and $ a+b\leq 16abc. $ Prove that
$$ a+b+kc^3\geq\sqrt[4]{\frac{4k} {27}}$$$$ a+b+kc^4\geq\frac{5} {8}\sqrt[5]{\frac{k} {2}}$$Where $ k>0. $
$$ a+b+3c^3\geq\sqrt{\frac{2} {3}}$$$$ a+b+2c^4\geq \frac{5} {8}$$
8 replies
sqing
Sunday at 12:46 PM
sqing
5 hours ago
A Collection of Good Problems from my end
SomeonecoolLovesMaths   10
N 5 hours ago by ReticulatedPython
This is a collection of good problems and my respective attempts to solve them. I would like to encourage everyone to post their solutions to these problems, if any. This will not only help others verify theirs but also perhaps bring forward a different approach to the problem. I will constantly try to update the pool of questions.

The difficulty level of these questions vary from AMC 10 to AIME. (Although the main pool of questions were prepared as a mock test for IOQM over the years)

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5
10 replies
SomeonecoolLovesMaths
Sunday at 8:16 AM
ReticulatedPython
5 hours ago
trapezoid
Darealzolt   0
Today at 2:03 AM
Let \(ABCD\) be a trapezoid such that \(A, B, C, D\) lie on a circle with center \(O\), and side \(AB\) is parallel to side \(CD\). Diagonals \(AC\) and \(BD\) intersect at point \(M\), and \(\angle AMD = 60^\circ\). It is given that \(MO = 10\). It is also known that the difference in length between \(AB\) and \(CD\) can be expressed in the form \(m\sqrt{n}\), where \(m\) and \(n\) are positive integers and \(n\) is square-free. Compute the value of \(m + n\).
0 replies
Darealzolt
Today at 2:03 AM
0 replies
Inequalities
sqing   2
N Today at 1:47 AM by sqing
Let $ a,b,c\geq 0 ,   2a +ab + 12a bc \geq 8. $ Prove that
$$  a+  (b+c)(a+1)+\frac{4}{5}  bc \geq 4$$$$  a+  (b+c)(a+0.9996)+ 0.77  bc \geq 4$$
2 replies
sqing
May 4, 2025
sqing
Today at 1:47 AM
anyone who can help me this 2 problems?
auroracliang   2
N Yesterday at 11:51 PM by ReticulatedPython
1. Let r be the radius of the largest circle which is tangent to the parabola y=x^2 at x=0 and which lies entirely on or inside (that is, above) the parabola, find r.

2. Counting number n has the following property,: if we take any 50 different numbers from 1,2,3, ... n, there always are two numbers with the difference of 7. what is the largest value among all value of n?


thanks a lot
2 replies
auroracliang
Nov 3, 2024
ReticulatedPython
Yesterday at 11:51 PM
What conic section is this? Is this even a conic section?
invincibleee   2
N Yesterday at 11:48 PM by ReticulatedPython
IMAGE

The points in this are given by
P = (sin2A, sin4A)∀A [0,2π]
Is this a conic section? what is this?
2 replies
invincibleee
Nov 15, 2024
ReticulatedPython
Yesterday at 11:48 PM
Spheres, ellipses, and cones
ReticulatedPython   0
Yesterday at 11:38 PM
A sphere is inscribed inside a cone with base radius $1$ and height $2.$ Another sphere of radius $r$ is internally tangent to the lateral surface of the cone, but does not intersect the larger inscribed sphere. A plane is tangent to both of these spheres, and passes through the inside of the cone. The intersection of the plane and the cone forms an ellipse. Find the maximum area of this ellipse.
0 replies
ReticulatedPython
Yesterday at 11:38 PM
0 replies
Looking for users and developers
derekli   13
N Yesterday at 11:31 PM by DreamineYT
Guys I've been working on a web app that lets you grind high school lvl math. There's AMCs, AIME, BMT, HMMT, SMT etc. Also, it's infinite practice so you can keep grinding without worrying about finding new problems. Please consider helping me out by testing and also consider joining our developer team! :P :blush:

Link: https://stellarlearning.app/competitive
13 replies
derekli
May 4, 2025
DreamineYT
Yesterday at 11:31 PM
trigonometric functions
VivaanKam   12
N Yesterday at 11:06 PM by aok
Hi could someone explain the basic trigonometric functions to me like sin, cos, tan etc.
Thank you!
12 replies
VivaanKam
Apr 29, 2025
aok
Yesterday at 11:06 PM
find number of elements in H
Darealzolt   1
N Yesterday at 6:47 PM by alexheinis
If \( H \) is the set of positive real solutions to the system
\[
x^3 + y^3 + z^3 = x + y + z
\]\[
x^2 + y^2 + z^2 = xyz
\]then find the number of elements in \( H \).
1 reply
Darealzolt
Yesterday at 1:50 AM
alexheinis
Yesterday at 6:47 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
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
aaravdodhia
2600 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
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
RedFireTruck
4221 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
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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 $\boxed{3-e}$. $\square$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
amitwa.exe
347 posts
#16 • 1 Y
Y by cubres
nicely done
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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 \[m=2154.\]
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
aman_maths
34 posts
#19 • 1 Y
Y by cubres
S3
This post has been edited 1 time. Last edited by aman_maths, May 25, 2024, 2:24 PM
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
amitwa.exe
347 posts
#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
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
amitwa.exe
347 posts
#22 • 1 Y
Y by cubres
Problem 10: Evaluate : $\frac{\displaystyle\sum_{k=1}^{n^2-1}\sqrt{n+\sqrt{k}}}{\displaystyle\sum_{k=1}^{n^2-1}\sqrt{n-\sqrt{k}}}$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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
$$\sum_{n=1}^{2020} na_n$$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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.
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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:
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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 $\boxed{2021!-1}$. $\square$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
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 $\Omega_1^{\Omega_2+\Omega_3}=\left (\frac{1}{2}\cdot\left (\frac{5}{16}\right )^3\right )^\frac{15}{8}=\boxed{\frac{5^\frac{45}{8}}{2^\frac{195}{8}}}$. $\square$
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
amitwa.exe
347 posts
#28 • 1 Y
Y by cubres
It is actually correct :10: . Nice solution
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
Aiden-1089
282 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
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
P162008
182 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
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
P162008
182 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
Z K Y
The post below has been deleted. Click to close.
This post has been deleted. Click here to see post.
soryn
5342 posts
#33
Y by
Very,very nice! Good job! Congratulations for all.. Instructivelu for me...
Z K Y
N Quick Reply
G
H
=
a