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Dhaka Secondary 2009/4

Posted: Fri Jan 21, 2011 6:14 pm
by BdMO
\[( 1-\frac{1}{2^2} ) ( 1- \frac{1}{3^2} ) ( 1- \frac{1}{4^2} ) ( 1- \frac{1}{5^2} ) (1- \frac{1}{6^2} ) \cdots (1- \frac{1}{1000^2} ) =?\]

Re: Dhaka Secondary 2009/4

Posted: Thu Jan 27, 2011 1:11 pm
by Zzzz
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? ;)

Re: Dhaka Secondary 2009/4

Posted: Thu Aug 25, 2011 2:23 am
by Shifat
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

Re: Dhaka Secondary 2009/4

Posted: Sat Feb 01, 2014 12:43 am
by Ridwan Abrar
What sequence did you find?