\[ P= \frac{2^2}{2^2 -1 } \times \frac{3^2}{3^2 - 1} \times \frac{4^2}{4^2 - 1}\times \cdot \cdot \cdot \times \frac{2006^2}{2006^2 - 1 }\]

\(P\) can Be Written as \(\frac{a}{b}\) Where \(a\) and \(b\) are coprime positive integers.

Find \(a+b\)

**Note**: This question was posed in the 2006 Harvard MIT Math Tournament

×

Problem Loading...

Note Loading...

Set Loading...