Mathematical Reflections 2016, Issue 1 - Problem S365

Problem:
Let $$a_k=\dfrac{(k^2+1)^2}{k^4+4}, \qquad k=1,2,3,\ldots.$$ Prove that for every positive integer $n$, $$a_1^na_2^{n-1}a_3^{n-2}\cdot \ldots \cdot a_n=\dfrac{2^{n+1}}{n^2+2n+2}.$$

 Proposed by Nguyen Viet Hung, Hanoi University of Science, Vietnam

Solution:
For any $k \in \mathbb{N}^*$, we have $$\renewcommand{\arraystretch}{2} \begin{array}{lll} a_k&=&\dfrac{(k^2+1)^2}{(k^2+2)^2-4k^2}\\&=&\dfrac{(k^2+1)^2}{(k^2-2k+2)(k^2+2k+2)}\\&=&\dfrac{k^2+1}{(k-1)^2+1}\cdot\dfrac{k^2+1}{(k+1)^2+1}. \end{array}$$
Let $P_j=\displaystyle \prod_{k=1}^j a_k$, where $j \in \mathbb{N}^*$. Then, $$P_j=\prod_{k=1}^j \dfrac{k^2+1}{(k-1)^2+1} \prod_{k=1}^j \dfrac{k^2+1}{(k+1)^2+1}=(j^2+1)\cdot\dfrac{2}{(j+1)^2+1}.$$ Therefore, we have
$$\renewcommand{\arraystretch}{2} \begin{array}{lll}\displaystyle a_1^na_2^{n-1}a_3^{n-2}\cdot \ldots \cdot a_n&=& \displaystyle \prod_{j=1}^n P_j\\ &=&\displaystyle\prod_{j=1}^n \dfrac{2(j^2+1)}{(j+1)^2+1}\\ &=& \displaystyle 2^n \prod_{j=1}^n \dfrac{j^2+1}{(j+1)^2+1}\\ &=&\displaystyle 2^n\cdot \dfrac{2}{(n+1)^2+1}\\&=&\displaystyle \dfrac{2^{n+1}}{n^2+2n+2}. \end{array}$$