AoPS Wiki talk:Problem of the Day/July 14, 2011

Problem

AoPSWiki:Problem of the Day/July 14, 2011

Solution

We begin by factoring the given expression, $\prod_{n=1}^{39}\frac{n^2+6n+9}{n^2+6n+8}$, to $\prod_{n=1}^{39}\frac{(n+3)^2}{(n+2)(n+4)}$. Then, writing this as multiple products and shifting the indices for clarity, we get $\frac{(\prod_{n=4}^{42}n)^2}{(\prod_{n=3}^{41}n)(\prod_{n=5}^{43}n)}$. Clearly, this equals $\frac{(\frac{42!}{6})^2}{(\frac{41!}{2})(\frac{43!}{24})}$. At this point, all that is left is arithmetic. The expression equals $\frac{(42!)^2*2*24}{41!*43!*6^2}=\frac{42!}{41!}*\frac{42!}{43!}*\frac{48}{36}=42*\frac{1}{43}*\frac{4}{3}$. This trivially simplifies to $\frac{168}{129}=\boxed{\frac{56}{43}}$.

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