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#1 h May 24, 2024, 8:52 AM • 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

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#2 h May 24, 2024, 8:52 AM • 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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#3 h May 24, 2024, 8:53 AM • 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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#4 h May 24, 2024, 8:53 AM • 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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#5 h May 24, 2024, 8:54 AM • 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

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#6 h May 24, 2024, 8:54 AM • 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.

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#7 h May 24, 2024, 8:55 AM • 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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#8 h May 24, 2024, 6:47 PM • 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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#9 h May 24, 2024, 9:54 PM • 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 $10,125/896$ ?

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#10 h May 25, 2024, 2:37 AM • 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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#11 h May 25, 2024, 2:40 AM • 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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#12 h May 25, 2024, 3:14 AM • 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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#13 h May 25, 2024, 3:15 AM • 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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#14 h May 25, 2024, 3:20 AM • 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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#15 h May 25, 2024, 4:36 AM • 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:

: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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#16 h May 25, 2024, 4:46 AM • 1 Y

Y by cubres

nicely done

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#17 h May 25, 2024, 6:15 AM • 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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#18 h May 25, 2024, 10:02 AM • 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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#19 h May 25, 2024, 2:23 PM • 1 Y

Y by cubres

S3

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#20 h May 25, 2024, 8:16 PM • 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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#21 h May 26, 2024, 9:35 AM • 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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#22 h May 26, 2024, 9:36 AM • 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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#23 h May 26, 2024, 9:36 AM • 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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#24 h May 26, 2024, 10:00 AM • 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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#25 h May 26, 2024, 10:25 AM • 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:

:blush:

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#26 h May 26, 2024, 2:46 PM • 2 Y

Y by cubres, soryn

amitwa.exe wrote:

i actually forgot to type $-f(2020)+f(2021)$ at the end :blush:

:blush:

Fortunately, the problem is still able to be solved :-D

:-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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#27 h May 26, 2024, 3:23 PM • 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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#28 h May 26, 2024, 4:09 PM • 1 Y

Y by cubres

It is actually correct :10:

:10:

. Nice solution

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#29 h May 26, 2024, 5:19 PM • 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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#31 h Apr 23, 2025, 4:01 AM • 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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#32 h Apr 23, 2025, 4:08 AM • 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

#33 h Apr 23, 2025, 6:25 AM

Y by

Very,very nice! Good job! Congratulations for all.. Instructivelu for me...

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