BdMO National Higher Secondary 2009/11

Discussion on Bangladesh Mathematical Olympiad (BdMO) National
User avatar
Moon
Site Admin
Posts:751
Joined:Tue Nov 02, 2010 7:52 pm
Location:Dhaka, Bangladesh
Contact:
BdMO National Higher Secondary 2009/11

Unread post by Moon » Sun Feb 06, 2011 11:35 pm

Problem 11:
Find $S$ where \
"Inspiration is needed in geometry, just as much as in poetry." -- Aleksandr Pushkin

Please install LaTeX fonts in your PC for better looking equations,
learn how to write equations, and don't forget to read Forum Guide and Rules.

User avatar
sm.joty
Posts:327
Joined:Thu Aug 18, 2011 12:42 am
Location:Dhaka

Re: BdMO National Higher Secondary 2009/11

Unread post by sm.joty » Sun Dec 18, 2011 4:32 pm

এর সমাধান কি ? :?:
এখানে কি দুটো Sum দিয়ে সমষ্টির সমষ্টি বোঝাইছে ?? confused :? :? :cry:
হার জিত চিরদিন থাকবেই
তবুও এগিয়ে যেতে হবে.........
বাধা-বিঘ্ন না পেরিয়ে
বড় হয়েছে কে কবে.........

User avatar
nafistiham
Posts:829
Joined:Mon Oct 17, 2011 3:56 pm
Location:24.758613,90.400161
Contact:

Re: BdMO National Higher Secondary 2009/11

Unread post by nafistiham » Sun Dec 18, 2011 7:09 pm

i think it is summation of summation
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
Using $L^AT_EX$ and following the rules of the forum are very easy but really important, too.Please co-operate.
Introduction:
Nafis Tiham
CSE Dept. SUST -HSC 14'
http://www.facebook.com/nafistiham
nafistiham@gmail

User avatar
amlansaha
Posts:100
Joined:Tue Feb 08, 2011 1:11 pm
Location:Khulna, Bangladesh
Contact:

Re: BdMO National Higher Secondary 2009/11

Unread post by amlansaha » Sun Dec 18, 2011 7:46 pm

যদিও সলুশন আমার অজানা, তবে এটা জানি যে সলুশনটা বেশ বড়।
অম্লান সাহা

sourav das
Posts:461
Joined:Wed Dec 15, 2010 10:05 am
Location:Dhaka
Contact:

Re: BdMO National Higher Secondary 2009/11

Unread post by sourav das » Thu Jan 23, 2014 2:14 pm

Looks like I still have some gun powder left in my gun :D .
Sketch Solution:
Given Statement
$=\sum^{\infty}_{m\neq n,m=1,n=1}\left ( \frac{m^2n}{3^m(m3^n+n3^m)}+ \frac{n^2m}{3^n(n3^m+m3^n)} \right )+ \sum^{\infty }_{m=1}\frac{m^3}{2.3^{2m}.m}$ (A tricky arrangement!)
$=\frac{1}{2}\left ( \sum^{\infty }_{m=1}(\frac{m}{3^m})^2+\sum^{\infty} _{m\neq n,m=1,n=1} 2\frac{mn}{3^m.3^n} \right )$

$=\frac{1}{2}\left ( \sum^{\infty }_{m=1}\frac{m}{3^m}\right )^2$.....(i)
Now Assume, $\sum^{\infty }_{m=1}\frac{m}{3^m}=S$
Then, you'll find out that, $3S= S+ \sum_{m=0}^{\infty }\frac{1}{3^m}$
But we know that, $\sum_{m=0}^{\infty }\frac{1}{3^m}=\frac{3}{2}$

So, $S=\frac{3}{4}$
Using (i), $Ans:\frac{1}{2}(\frac{3}{4})^2=\frac{9}{32}$
You spin my head right round right round,
When you go down, when you go down down......
(-$from$ "$THE$ $UGLY$ $TRUTH$" )

User avatar
*Mahi*
Posts:1175
Joined:Wed Dec 29, 2010 12:46 pm
Location:23.786228,90.354974
Contact:

Re: BdMO National Higher Secondary 2009/11

Unread post by *Mahi* » Thu Jan 23, 2014 2:40 pm

Given \
\[= \sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\dfrac{1}{\frac{3^m}{m}(\frac{3^m}{m}+\frac{3^n}{n})}\]
By symmetry \[S = \sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\dfrac{1}{\frac{3^n}{n}(\frac{3^m}{m}+\frac{3^n}{n})} \]
So, \[2S = \sum_{m=1}^{\infty}\sum_{n=1}^{\infty} \left (\dfrac{1}{\frac{3^n}{n}(\frac{3^m}{m}+\frac{3^n}{n})} + \dfrac{1}{\frac{3^m}{m}(\frac{3^m}{m}+\frac{3^n}{n})} \right )\]
\[= \sum_{m=1}^{\infty}\sum_{n=1}^{\infty} \dfrac{ \frac{3^m}m +\frac {3^n}{n}} { \frac {3^m}{m} \frac {3^n}{n} (\frac{3^m}{m}+\frac{3^n}{n})}\]
\[= \sum_{m=1}^{\infty}\sum_{n=1}^{\infty} \dfrac {1} {\frac {3^m}{m} \frac {3^n}{n} }\]
\[= \sum_{m=1}^{\infty}\sum_{n=1}^{\infty} \frac{m}{3^m} \frac n{3^n}\]
\[ =\sum_{m=1}^{\infty} \frac{m}{3^m} \left ( \sum_{n=1}^{\infty} \frac n{3^n} \right ) \]
\[ = \left (\sum_{m=1}^{\infty} \frac{m}{3^m} \right) \left ( \sum_{n=1}^{\infty} \frac n{3^n} \right ) \]
Now, \[ \left (\sum_{m=1}^{\infty} \frac{m}{3^m} \right) = \frac 3 4 \]
So, \[2S = \frac 3 4 \times \frac 3 4 = \frac 9 {16}\]
And thus \
As far as I remember, the problem setter was Abir vai. He showed it in 2010 winter camp too.

[hide]Edit:
ভাত খাইতে গিয়া দেরি হয়া গেল দেখি -_-[/hide]
Please read Forum Guide and Rules before you post.

Use $L^AT_EX$, It makes our work a lot easier!

Nur Muhammad Shafiullah | Mahi

User avatar
Masum
Posts:592
Joined:Tue Dec 07, 2010 1:12 pm
Location:Dhaka,Bangladesh

Re: BdMO National Higher Secondary 2009/11

Unread post by Masum » Thu Jan 23, 2014 5:04 pm

I found a straight solution without even symmetry, and I think this solutions makes more sense than the symmetry one. $S$ can be re-written as
\[\sum_{n=1}^\infty\sum_{n=1}^\infty \dfrac{n^2n}{3^n\cdot(n3^n+n3^n)}=\dfrac12\sum_{n=1}^\infty\sum_{n=1}^\infty \dfrac{n^2}{\left(3^n\right)^2}=\dfrac12\sum_{n=1}^\infty\sum_{n=1}^\infty \left(\dfrac{n}{3^n}\right)\left(\dfrac{n}{3^n}\right)=\dfrac12\left(\sum_{n=1}^\infty\dfrac n{3^n}\right)^2\]
Note the trick to write the sum as the last portion. It is actually the following:
\[\sum_{i=1}^k\sum_{j=1}^la_ib_j=\left(\sum_{i=1}^ka_i\right)\left(\sum_{j=1}^lb_i\right)\]
Now $T=\sum\limits_{n=1}^\infty\dfrac n{3^n}$ which can be found easily.
One one thing is neutral in the universe, that is $0$.

User avatar
*Mahi*
Posts:1175
Joined:Wed Dec 29, 2010 12:46 pm
Location:23.786228,90.354974
Contact:

Re: BdMO National Higher Secondary 2009/11

Unread post by *Mahi* » Thu Jan 23, 2014 8:55 pm

Masum wrote:I found a straight solution without even symmetry, and I think this solutions makes more sense than the symmetry one. $S$ can be re-written as
\[\sum_{n=1}^\infty\sum_{n=1}^\infty \dfrac{n^2n}{3^n\cdot(n3^n+n3^n)}=\dfrac12\sum_{n=1}^\infty\sum_{n=1}^\infty \dfrac{n^2}{\left(3^n\right)^2}=\dfrac12\sum_{n=1}^\infty\sum_{n=1}^\infty \left(\dfrac{n}{3^n}\right)\left(\dfrac{n}{3^n}\right)=\dfrac12\left(\sum_{n=1}^\infty\dfrac n{3^n}\right)^2\]
Note the trick to write the sum as the last portion. It is actually the following:
\[\sum_{i=1}^k\sum_{j=1}^la_ib_j=\left(\sum_{i=1}^ka_i\right)\left(\sum_{j=1}^lb_i\right)\]
Now $T=\sum\limits_{n=1}^\infty\dfrac n{3^n}$ which can be found easily.
I am not quite sure about this :/
For example, let $\{a_i \} = \{i\}$ and $\{b_i \} = \{\frac 1 i\}$.
Then according to your method \[\sum _{i=1}^{n} \sum _{j=1}^{n} a_ib_j = \sum _{i=1}^{n} \sum _{i=1}^{n} a_ib_i = \sum _{i=1}^{n} \sum _{i=1}^{n} 1 = \sum _{i=1}^{n} \sum _{i=1}^{n} 1 \cdot 1 = \left (\sum _{i=1}^{n} 1 \right)^2 = n^2 \]
While \[\sum _{i=1}^{n} \sum _{j=1}^{n} a_ib_j = \left( \sum _{i=1}^{n} i \right) \left( \sum _{i=1}^{n} \frac 1 i \right) \] which is definitely not $n^2$ :|
Please read Forum Guide and Rules before you post.

Use $L^AT_EX$, It makes our work a lot easier!

Nur Muhammad Shafiullah | Mahi

User avatar
Masum
Posts:592
Joined:Tue Dec 07, 2010 1:12 pm
Location:Dhaka,Bangladesh

Re: BdMO National Higher Secondary 2009/11

Unread post by Masum » Fri Jan 24, 2014 9:25 am

Hmm. May be a miss-statement. This is infact a generalization of $a(c+d)+b(c+d)=(a+b)(c+d)$ so far as I am concerned.
One one thing is neutral in the universe, that is $0$.

User avatar
*Mahi*
Posts:1175
Joined:Wed Dec 29, 2010 12:46 pm
Location:23.786228,90.354974
Contact:

Re: BdMO National Higher Secondary 2009/11

Unread post by *Mahi* » Fri Jan 24, 2014 1:13 pm

Masum wrote:Hmm. May be a miss-statement. This is infact a generalization of $a(c+d)+b(c+d)=(a+b)(c+d)$ so far as I am concerned.
$\sum _{i=1}^{n} \sum _{j=1}^{n} a_ib_j = (\sum _{i=1}^{n} a_i) (\sum _{j=1}^{n} b_j)$ might be, but I am not sure about $\sum _{i=1}^{n} \sum _{j=1}^{n} a_ib_j = \sum _{i=1}^{n} \sum _{i=1}^{n} a_ib_i $
Please read Forum Guide and Rules before you post.

Use $L^AT_EX$, It makes our work a lot easier!

Nur Muhammad Shafiullah | Mahi

Post Reply