Question
The value of $\frac{\left( 1^{4} + \frac{1}{4} \right)\left( 3^{4} + \frac{1}{4} \right)\cdots\cdots\left( (2n - 1)^{4} + \frac{1}{4} \right)}{\ }$ is equal to
$$\left( 2^{4} + \frac{1}{4} \right)\left( 4^{4} + \frac{1}{4} \right)\cdots\cdots\left( (2n)^{4} + \frac{1}{4} \right)$$
Options
Solution
Sol. [Key Concept: Use $k^{4 +}\frac{14 = (}{k}^{2 -}k + \frac{12)(}{k}^{2 +}k + \frac{12)}{}$ to telescope the product/ratio.]
$$k^{4} + \frac{1}{4} = \left( k^{2} - k + \frac{1}{2} \right)\left( k^{2} + k + \frac{1}{2} \right)$$
$$\prod_{k = 1}^{n}\mspace{2mu}\frac{\left( (2k - 1)^{2} - (2k - 1) + \frac{1}{2} \right)\left( (2k - 1)^{2} + (2k - 1) + \frac{1}{2} \right)}{\left( (2k)^{2} - 2k + \frac{1}{2} \right)\left( (2k)^{2} + 2k + \frac{1}{2} \right)}$$
$$\begin{matrix} & \ = \prod_{k = 1}^{n}\mspace{2mu}\mspace{2mu}\frac{(2k - 1)^{2} - (2k - 1) + \frac{1}{2}}{(2k)^{2} + 2k + \frac{1}{2}} \\ & \ = \frac{\frac{1}{2}}{4n^{2} + 2n + \frac{1}{2}} = \frac{1}{8n^{2} + 4n + 1} \end{matrix}$$
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