<s>[INST] A question on summation of series... 
If $$\frac{\left(1^4+\frac14\right)\left(3^4+\frac14\right)\ldots\left((2n-1)^4+\frac14\right)}{\left(2^4+\frac14\right)\left(4^4+\frac14\right)\ldots\left((2n)^4+\frac14\right)}=\frac1{k_1n^2+k_2n+k_3}$$ then $k_1-k_2+k_3$ is equal to . . . ?

What I  basically tried was simply putting the value of $n$ from $1$ to $3$. By doing so I got $3$ simultaneous equations and hence I calculated the desired answer.
What I want to know is that is there any other elegant way to solve the above problem  because my process is very time consuming ..looking forward to here from my friends thanks in advance......
 [/INST] Use the Sophie-Germain identity: $$a^4+4b^4=(a^2+2b^2)^2-(2ab)^2=(a^2-2ab+2b^2)(a^2+2ab+2b^2)$$Using this, we can factor an expression of the form $m^4+\frac{1}{4}$ into $$m^4+4\bigg(\frac{1}{2}\bigg)^4=(m^2-m+\frac{1}{2})(m^2+m+\frac{1}{2})$$Therefore, the LHS of your equation is equal to $$\frac{\displaystyle\prod_{i=1}^{n}\bigg((2i-1)^2-(2i-1)+\frac{1}{2}\bigg)\bigg((2i-1)^2+(2i-1)+\frac{1}{2}\bigg)}{\displaystyle\prod_{i=1}^{n}\bigg((2i)^2-(2i)+\frac{1}{2}\bigg)\bigg((2i)^2+(2i)+\frac{1}{2}\bigg)}$$Note that $$(2i-1)^2+(2i-1)+\dfrac{1}{2}=(2i)^2-4i+1+2i-1+\dfrac{1}{2}=(2i)^2-(2i)+\dfrac{1}{2}$$and $$(2i-2)^2+(2i-2)+\frac{1}{2}=(2i)^2-8i+4+2i-2+\frac{1}{2}=(2i-1)^2-(2i-1)+\frac{1}{2}$$So after cancelling, our expression above is just $$\frac{(2-1)^2-(2-1)+\frac{1}{2}}{(2n)^2+(2n)+\frac{1}{2}}=\frac{1}{8n^2+4n+1}$$and the answer is $8-4+1=5$.
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