BdMO National Higher Secondary 2009/11

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BdMO National Higher Secondary 2009/11

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

Problem 11:
Find $S$ where \
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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:
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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
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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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

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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), $Ans:\frac{1}{2}(\frac{3}{4})^2=\frac{9}{32}$
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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.

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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.
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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 $n^2$ :|
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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.
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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 $
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