An algebra problem by MaFiA maNiAc

Algebra Level 3

$\large \left( 1 - \dfrac1{2^2} \right)\left( 1 - \dfrac1{3^2} \right)\left( 1 - \dfrac1{4^2} \right)\cdots \left( 1 - \dfrac1{2007^2} \right) = \, ?$

\frac{1004}{2007}

\frac{2007}{2008}

1

\frac{2008}{2007}

2 solutions

Chew-Seong Cheong
Sep 7, 2016

$\begin{aligned} P&= \left(1-\frac 1 {2^2} \right) \left(1-\frac 1 {3^2} \right) \left(1-\frac 1 {4^2} \right)... \left(1-\frac 1 {2007^2} \right) \\ & =\prod_{n=2}^{2007} \left(1-\frac 1 {n^2} \right) \\ & =\prod_{n=2}^{2007} \frac {n^2 - 1} {n^2} \\ & =\prod_{n=2}^{2007} \frac {(n- 1) (n+1)} {n^2} \\ & = \frac {2006! \cdot 2008!}{2 \cdot (2007!)^2} \\ & = \boxed {\dfrac {1004}{2007}} \end{aligned}$

Suresh Jh
May 6, 2018

An algebra problem by MaFiA maNiAc

2 solutions

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