Summation Or Something Else

Relevant wiki: Sophie Germain Identity

$\begin{aligned} Q & = \frac {\prod_{k=1}^n \left((2k-1)^4+\frac 14\right)}{\prod_{k=1}^n \left((2k)^4+\frac 14\right)} & \small \color{#3D99F6} \text{Multiply up and down by }4^n \\ & = \frac {\prod_{k=1}^n \left(4(2k-1)^4+1\right)}{\prod_{k=1}^n \left(4(2k)^4+1\right)} & \small \color{#3D99F6} \text{By Sophie Germain's identity: } \\ & = \frac {\prod_{k=1}^n \cancel{\left(8k^2-4k+1\right)}\left(8k^2-12k+5\right)}{\prod_{k=1}^n \left(8k^2+4k+1\right)\cancel{\left(8k^2-4k+1\right)}} & \small \color{#3D99F6} a^4+4b^4 = (a^2+2ab+ 2b^2)(a^2-2ab+2b^2) \\ & = \frac {\prod_{k=1}^n \left(8k^2-12k+5\right)}{\prod_{k=1}^n \left(8k^2+4k+1\right)} \\ & = \frac {\overbrace{(8(1^2)-12(1)+5)}^{\color{#3D99F6}k=1} \prod_{\color{#3D99F6}k=2}^{\color{#3D99F6}n} \left(8{\color{#3D99F6}k^2}-12{\color{#3D99F6}k}+5\right)}{\prod_{k=1}^n \left(8k^2+4k+1\right)} \\ & = \frac {{\color{#3D99F6}(1)}\prod_{\color{#D61F06}k=1}^{\color{#D61F06}n-1} \left(8{\color{#D61F06}(k+1)^2}-12{\color{#D61F06}(k+1)}+5\right)}{\prod_{k=1}^n \left(8k^2+4k+1\right)} \\ & = \frac {\prod_{k=1}^{n-1} \left(8k^2+4k+1\right)}{\prod_{k=1}^n \left(8k^2+4k+1\right)} \\ & = \frac 1{8n^2+4n+1} \end{aligned}$

Therefore, $a+b+c = 8+4+1=\boxed{13}$ .

$\begin{aligned} & P =\displaystyle\prod_{n=1}^{m}\left( \dfrac{(2n-1)^4+\frac{1}{4}}{(2n)^4 +\frac{1}{4}}\right) \\& P = \displaystyle\prod_{n=1}^{m} \left(\dfrac{4(2n-1)^4 +1}{4(2n)^4 +1}\right)\\& P = \displaystyle\prod_{n=1}^{m} \left(\dfrac{4q^4 +p^4}{4b^4 +a^4}\right)\end{aligned}$ where $a =1 , b = 2n, p = 1, q = 2n-1$ and using Sophie German Identity for the expansion of $\begin{aligned} & \dfrac{p^4 +4q^4}{a^4 +4b^4 } = \dfrac{((p+q)^2+q^2)((p-q)^2+q^2)}{((a+b)^2+b^2)((a-b)^2+b^2))}\\& \dfrac{p^4+4q^4}{a^4+4b^4} =\dfrac{((1+2n-1)^2 +(2n-1)^2)((1-(2n-1))^2+(2n-1)^2)}{((1+2n)^2+(2n)^2)((1-2n)^2+(2n)^2)} \\& \dfrac{p^4+4q^4}{a^4+4b^4} = \dfrac{(2n)^2 +(2n-1)^2)((2-2n)^2 + (2n-1)^2)}{(1+2n)^2+(2n)^2)((1-2n)^2 + (2n)^2)} \\& \dfrac{p^4+4q^4}{a^4+4b^4} = \dfrac{(2-2n)^2+(2n-1)^2}{(2n+1)^2 +(2n)^2} =\dfrac{4-8n +4n^2 +4n^2 -4n+1}{4n^2+4n+1 +4n^2} =\dfrac{8n^2 -12n +5}{8n^2 + 4n+1} \end{aligned}$ $\therefore \displaystyle\prod_{n=1}^{m} \dfrac{(2n-1)^4 +1}{(2n)^4+1} = \displaystyle\prod_{n=1}^{m} \dfrac{8n^2-12n+4}{8n^2 +4n+1} = \dfrac{1}{\boxed{8m^2+4m+1}}$ Note $\displaystyle\prod_{n=1}^{m} \dfrac{8n^2-12n+5}{8n^2+4n+1} = \dfrac{1}{13}\times \dfrac{13}{41} \times \dfrac{41}{85}\times \cdots \times\dfrac{}{8m^2-12m+5}\times \dfrac{8m^2-12m+5}{8m^2+4m+1}$ shows that the above product is telescoping product.