2006 Sum

Algebra Level 3

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

$P$ 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

2 solutions

Siddhartha Srivastava
Apr 17, 2014

Define $S = \dfrac{2}{2-1} \times \dfrac{3}{3-1} \times \dots \times \dfrac{2005}{2005 - 1} \times \dfrac{2006}{2006 - 1}$

And $R = \dfrac{2}{2+1} \times \dfrac{3}{3+1} \times \dots \times \dfrac{2005}{2005 + 1} \times \dfrac{2006}{2006 + 1}$

Now, Note that $S \times R = P$

Now, $S = \dfrac{2}{1} \times \dfrac{3}{2} \times \dots \times \dfrac{2005}{2004} \times \dfrac{2006}{2005}$

$\Rightarrow S = \dfrac{2006}{1}$

And $R = \dfrac{2}{3} \times \dfrac{3}{4} \times \dots \times \dfrac{2005}{2006} \times \dfrac{2006}{2007}$

$\Rightarrow R = \dfrac{2}{2007}$

$P = R \times S = 2006 \times \dfrac{2}{2007} = \dfrac{4012}{2007}$

Therefore $a = 4012, b = 2007$ and $a+b = \boxed{6019}$

\displaystyle P= \frac{2 *2}{1*3 } \times \frac{3*3}{2*4} \times \frac{4*4}{3*5}\times \cdot \cdot \cdot \cdot \cdot \cdot \cdot \times \frac{2004*2004}{2003*2005 } \times \frac{2005*2005 }{2004*2006} \times \frac{2006*2006}{2005*2007 }

So We Have the Product of Extremes

$\displaystyle P= \frac{2 }{1} \times \frac{2006}{2007} =$ $\displaystyle \boxed{\frac{4012 }{2007}}$

Gabriel Merces - 7 years, 1 month ago

Nice Solution. I felt splitting the product up would make the solution more clean and easier to understand.

Siddhartha Srivastava - 7 years, 1 month ago

Extremely Great Solution... Crisp Sweet and easy.... Keep it up Well Doe!

Mehul Arora - 6 years, 7 months ago

Kartik Sharma
May 1, 2014

is it really a Harvard MIT Math Tournament question?

MAYBE ...................................................

math man - 6 years, 8 months ago