Expression Simplification 5
Let $ n $ be a positive integer. Simplify the expression
\[\frac{(2^4 + \frac{1}{4})(4^4 + \frac{1}{4}) \dotsm [(2n)^4 + \frac{1}{4}]}{(1^4 + \frac{1}{4})(3^4 + \frac{1}{4}) \dotsm [(2n - 1)^4 + \frac{1}{4}]}.\]
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- $\frac{a}{b}$
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- $a^n$
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Solution
Let
\[f(m) = m^4 + \frac{1}{4} = \frac{4m^4 + 1}{4}.\]We can factor this with a little give and take:
\begin{align*}
f(m) &= \frac{4m^4 + 1}{4} \\
&= \frac{4m^4 + 4m^2 + 1 - 4m^2}{4} \\
&= \frac{(2m^2 + 1)^2 - (2m)^2}{4} \\
&= \frac{(2m^2 + 2m + 1)(2m^2 - 2m + 1)}{4}.\end{align*}Now, let $ g(m) = 2m^2 + 2m + 1 $. Then
\[g(m - 1) = 2(m - 1)^2 + 2(m - 1) + 1 = 2m^2 - 2m + 1.\]Hence,
\[f(m) = \frac{g(m) g(m - 1)}{4}.\]Therefore,
\begin{align*}
\frac{(2^4 + \frac{1}{4})(4^4 + \frac{1}{4}) \dotsm [(2n)^4 + \frac{1}{4}]}{(1^4 + \frac{1}{4})(3^4 + \frac{1}{4}) \dotsm [(2n - 1)^4 + \frac{1}{4}]} &= \frac{f(2) f(4) \dotsm f(2n)}{f(1) f(3) \dotsm f(2n - 1)} \\
&= \frac{\frac{g(2) g(1)}{4} \cdot \frac{g(4) g(3)}{4} \dotsm \frac{g(2n) g(2n - 1)}{4}}{\frac{g(1) g(0)}{4} \cdot \frac{g(3) g(2)}{4} \dotsm \frac{g(2n - 1) g(2n - 2)}{4}} \\
&= \frac{g(2n)}{g(0)} \\
&= 2(2n)^2 + 2(2n) + 1 \\
&= \boxed{8n^2 + 4n + 1}.\end{align*}
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