Dhaka Secondary 2009/4

Problem for Secondary Group from Divisional Mathematical Olympiad will be solved here.
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Please don't post problems (by starting a topic) in the "Secondary: Solved" forum. This forum is only for showcasing the problems for the convenience of the users. You can post the problems in the main Divisional Math Olympiad forum. Later we shall move that topic with proper formatting, and post in the resource section.
BdMO
Posts: 134
Joined: Tue Jan 18, 2011 1:31 pm

Dhaka Secondary 2009/4

Unread post by BdMO » Fri Jan 21, 2011 6:14 pm

\[( 1-\frac{1}{2^2} ) ( 1- \frac{1}{3^2} ) ( 1- \frac{1}{4^2} ) ( 1- \frac{1}{5^2} ) (1- \frac{1}{6^2} ) \cdots (1- \frac{1}{1000^2} ) =?\]

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Zzzz
Posts: 172
Joined: Tue Dec 07, 2010 6:28 am
Location: 22° 48' 0" N / 89° 33' 0" E

Re: Dhaka Secondary 2009/4

Unread post by Zzzz » Thu Jan 27, 2011 1:11 pm

Nice problem ..
$(1- \frac {1}{n^2})= \frac {(n+1)(n-1)}{n^2}$ Now where can we find a $(n-1)$ and a $(n+1)$ as denominator? ;)
Every logical solution to a problem has its own beauty.
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Shifat
Posts: 53
Joined: Sun Jul 31, 2011 12:21 pm
Location: Dhaka, Bangladesh

Re: Dhaka Secondary 2009/4

Unread post by Shifat » Thu Aug 25, 2011 2:23 am

I found a sequence out of the serial multiplications, my result is $\frac{1001}{2000}$
I just observed the multiplications of first 7 numbers, please check it though

Ridwan Abrar
Posts: 13
Joined: Mon Aug 05, 2013 8:01 pm

Re: Dhaka Secondary 2009/4

Unread post by Ridwan Abrar » Sat Feb 01, 2014 12:43 am

What sequence did you find?

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