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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.

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0 replies
jlacosta
May 1, 2025
0 replies
Maximum number of terms in the sequence
orl   11
N 6 minutes ago by navier3072
Source: IMO LongList, Vietnam 1, IMO 1977, Day 1, Problem 2
In a finite sequence of real numbers the sum of any seven successive terms is negative and the sum of any eleven successive terms is positive. Determine the maximum number of terms in the sequence.
11 replies
orl
Nov 12, 2005
navier3072
6 minutes ago
Combinatorics
P162008   2
N 15 minutes ago by cazanova19921
Let $m,n \in \mathbb{N}.$ Let $[n]$ denote the set of natural numbers less than or equal to $n.$

Let $f(m,n) = \sum_{(x_1,x_2,x_3, \cdots, x_m) \in [n]^{m}} \frac{x_1}{x_1 + x_2 + x_3 + \cdots + x_m} \binom{n}{x_1} \binom{n}{x_2} \binom{n}{x_3} \cdots \binom{n}{x_m} 2^{\left(\sum_{i=1}^{m} x_i\right)}$

Compute the sum of the digits of $f(4,4).$
2 replies
P162008
an hour ago
cazanova19921
15 minutes ago
USAMO 2003 Problem 1
MithsApprentice   68
N 16 minutes ago by Mamadi
Prove that for every positive integer $n$ there exists an $n$-digit number divisible by $5^n$ all of whose digits are odd.
68 replies
MithsApprentice
Sep 27, 2005
Mamadi
16 minutes ago
Centroid, altitudes and medians, and concyclic points
BR1F1SZ   3
N 42 minutes ago by EeEeRUT
Source: Austria National MO Part 1 Problem 2
Let $\triangle{ABC}$ be an acute triangle with $BC > AC$. Let $S$ be the centroid of triangle $ABC$ and let $F$ be the foot of the perpendicular from $C$ to side $AB$. The median $CS$ intersects the circumcircle $\gamma$ of triangle $\triangle{ABC}$ at a second point $P$. Let $M$ be the point where $CS$ intersects $AB$. The line $SF$ intersects the circle $\gamma$ at a point $Q$, such that $F$ lies between $S$ and $Q$. Prove that the points $M,P,Q$ and $F$ lie on a circle.

(Karl Czakler)
3 replies
BR1F1SZ
Monday at 9:45 PM
EeEeRUT
42 minutes ago
Geometry Proof
strongstephen   10
N 3 hours ago by martianrunner
Proof that choosing four distinct points at random has an equal probability of getting a convex quadrilateral vs a concave one.
not cohesive proof alert!

NOTE: By choosing four distinct points, that means no three points lie on the same line on the Gaussian Plane.
NOTE: The probability of each point getting chosen don’t need to be uniform (as long as it is symmetric about the origin), you just need a way to choose points in the infinite plane (such as a normal distribution)

Start by picking three of the four points. Next, graph the regions where the fourth point would make the quadrilateral convex or concave. In diagram 1 below, you can see the regions where the fourth point would be convex or concave. Of course, there is the centre region (the shaded triangle), but in an infinite plane, the probability the fourth point ends up in the finite region approaches 0.

Next, I want to prove to you the area of convex/concave, or rather, the probability a point ends up in each area, is the same. Referring to the second diagram, you can flip each concave region over the line perpendicular to the angle bisector of which the region is defined. (Just look at it and you'll get what it means.) Now, each concave region has an almost perfect 1:1 probability correspondence to another convex region. The only difference is the finite region (the triangle, shaded). Again, however, the actual significance (probability) of this approaches 0.

If I call each of the convex region's probability P(a), P(c), and P(e) and the concave ones P(b), P(d), P(f), assuming areas a and b are on opposite sides (same with c and d, e and f) you can get:
P(a) = P(b)
P(c) = P(d)
P(e) = P(f)

and P(a) + P(c) + P(e) = P(convex)
and P(b) + P(d) + P(f) = P(concave)

therefore:
P(convex) = P(concave)
10 replies
strongstephen
Yesterday at 4:54 AM
martianrunner
3 hours ago
A Collection of Good Problems from my end
SomeonecoolLovesMaths   24
N 6 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
24 replies
SomeonecoolLovesMaths
May 4, 2025
ReticulatedPython
6 hours ago
n and n+100 have odd number of divisors (1995 Belarus MO Category D P2)
jasperE3   4
N Yesterday at 9:50 PM by KTYC
Find all positive integers $n$ so that both $n$ and $n + 100$ have odd numbers of divisors.
4 replies
jasperE3
Apr 6, 2021
KTYC
Yesterday at 9:50 PM
Closed form expression of 0.123456789101112....
ReticulatedPython   3
N Yesterday at 8:15 PM by ReticulatedPython
Is there a closed form expression for the decimal number $$0.123456789101112131415161718192021...$$which is defined as all the natural numbers listed in order, side by side, behind a decimal point, without commas? If so, what is it?
3 replies
ReticulatedPython
Yesterday at 8:05 PM
ReticulatedPython
Yesterday at 8:15 PM
primes and perfect squares
Bummer12345   5
N Yesterday at 8:08 PM by Shan3t
If $p$ and $q$ are primes, then can $2^p + 5^q + pq$ be a perfect square?
5 replies
Bummer12345
Monday at 5:08 PM
Shan3t
Yesterday at 8:08 PM
trapezoid
Darealzolt   1
N Yesterday at 7:38 PM by vanstraelen
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 \(m + n\).
1 reply
Darealzolt
Yesterday at 2:03 AM
vanstraelen
Yesterday at 7:38 PM
Polynomial Minimization
ReticulatedPython   1
N Yesterday at 5:36 PM by clarkculus
Consider the polynomial $$p(x)=x^{n+1}-x^{n}$$, where $x, n \in \mathbb{R+}.$

(a) Prove that the minimum value of $p(x)$ always occurs at $x=\frac{n}{n+1}.$
1 reply
ReticulatedPython
Yesterday at 5:07 PM
clarkculus
Yesterday at 5:36 PM
Easy one
irregular22104   0
Yesterday at 5:03 PM
Given two positive integers a,b written on the board. We apply the following rule: At each step, we will add all the numbers that are the sum of the two numbers on the board so that the sum does not appear on the board. For example, if the two initial numbers are 2.5, then the numbers on the board after step 1 are 2,5,7; after step 2 are 2,5,7,9,12;...
1) With a = 3; b = 12, prove that the number 2024 cannot appear on the board.
2) With a = 2; b = 3, prove that the number 2024 can appear on the board.
0 replies
irregular22104
Yesterday at 5:03 PM
0 replies
This shouldn't be a problem 15
derekli   2
N Yesterday at 4:09 PM by aarush.rachak11
Hey guys I was practicing AIME and came across this problem which is definitely misplaced. It asks for the surface area of a plane within a cylinder which we can easily find out using a projection that is easy to find. I think this should be placed in problem 10 or below. What do you guys think?
2 replies
derekli
Yesterday at 2:15 PM
aarush.rachak11
Yesterday at 4:09 PM
Regular tetrahedron
vanstraelen   7
N Yesterday at 3:46 PM by ReticulatedPython
Given the points $O(0,0,0),A(1,0,0),B(\frac{1}{2},\frac{\sqrt{3}}{2},0)$
a) Determine the point $C$, above the xy-plane, such that the pyramid $OABC$ is a regular tetrahedron.
b) Calculate the volume.
c) Calculate the radius of the inscribed sphere and the radius of the circumscribed sphere.
7 replies
vanstraelen
May 4, 2025
ReticulatedPython
Yesterday at 3:46 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
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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
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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
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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
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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
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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
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
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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
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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
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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 $a$)

$S=\sum_{k=1}^{2018} \left\lfloor{\frac{103k}{2019}}\right\rfloor=\sum_{k=1}^{2018} \frac{103k}{2019}-\sum_{k=1}^{2018} \{\frac{103k}{2019}\}=\frac{103}{2019}\frac{2018\cdot 2019}{2}-\sum_{k=1}^{2018} \frac{k}{2019}=103\cdot 1009-\frac{1}{2019}\frac{2018\cdot2019}{2}=102\cdot1009=\boxed{102918}$
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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
This post has been edited 1 time. Last edited by aman_maths, May 25, 2024, 2:24 PM
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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
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
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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
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
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P162008
183 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
183 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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