### Problem Evaluate $ \prod_{n=1}^{2004} \frac{n^{2}+2 n-1}{n^{2}+n+\sqrt{2}-2} $ ##### Hints - It is unlikely to be expected to do more than 2000 calculations, so it is likely that this product is [[telescoping]]. #### Solution This problem is best solved by starting with some elegant factoring on the argument of the product. We will start by re-writing the numerator and denominator $ \prod_{n=1}^{2004} \frac{n^{2}+2 n+1-2}{n^{2}-2+n+\sqrt{2}} $ Now the factoring begins. - in the numerator, the square trinomial is factored. - in the denominator, the difference of squares is factored $ \prod_{n=1}^{2004} \frac{(n+1)^2-2}{(n-\sqrt2)(n+\sqrt2)+n+\sqrt{2}} $ The factoring is not over yet - in the numerator, the new difference of squares is factored. - in the denominator, extracting the common factor leaves behind a factor shared with the numerator $ \prod_{n=1}^{2004} \frac{\cancel{(n+1-\sqrt2)}(n+1+\sqrt2)}{\cancel{(n-\sqrt2+1)}(n+\sqrt2)} $ Now the product's telescoping nature is revealed, and only the denominator for $n=1$ and the numerator for $n=2004$ will remain. $\begin{aligned}\frac{\cancel{2+\sqrt2}}{1+\sqrt2}\cdot\frac{\cancel{3+\sqrt2}}{\cancel{2+\sqrt2}}\cdots\frac{2005+\sqrt2}{\cancel{2004+\sqrt2}}=&\\\frac{2005+\sqrt2}{1+\sqrt2}=&\\\frac{(2005+\sqrt2)(\sqrt2-1)}{(1+\sqrt2)(\sqrt2-1)}=&\\\frac{2005\sqrt2-2005+2-\sqrt2}{2-1}=&\\\boxed{2004\sqrt2-2003}=&\end{aligned}$