<s>[INST] sum of 100 terms of logarithmic expression 
Calculate value of


$\displaystyle \sum^{100}_{k=1}\ln\bigg(\frac{(2k+1)^4+\frac{1}{4}}{16k^4+\frac{1}{4}}\bigg)$

My try :: $\displaystyle x^4+4y^4$
$=(x^2+2xy+2y^2)(x^2-2xy+2y^2)$
So sum $\displaystyle \sum^{100}_{k=1}\ln\bigg(\frac{1+4(2k+1)^4}{1+4(2k)^4}\bigg)$
$\displaystyle =\sum^{100}_{k=1}\ln\bigg[\frac{(1+2(2k+1)+2(2k+1)^2)(1-2(2k+1)+2(2k+1)^2)}{(1+2(2k)+2(2k)^2)(1-2(2k)+2(2k)^2)}\bigg]$
How can I decompose that complex expression into partial fractions?
 [/INST] $$g(k)=\dfrac{(2k+1)^4+\dfrac14}{(2k)^4+\dfrac14}=\dfrac{(4k+2)^4+4\cdot1^4}{(4k)^4+4\cdot1^4}$$
Using your formula only,
$\displaystyle x^4+4y^4=(x^2+2y^2)^2-(2xy)^2=(x^2+2xy+2y^2)(x^2-2xy+2y^2)$
$x=4k+2, y=1$
$\displaystyle\implies x^2+2xy+2y^2=(4k+2)^2+2(4k+2)+2=16k^2+24k+10\  \ \ \ (1)$
$x^2-2xy+2y^2=(4k+2)^2-2(4k+2)+2=16k^2+8k+2\  \ \ \ (2)$
$x=4k, y=1$
$\displaystyle\implies x^2+2xy+2y^2=(4k)^2+2(4k)+2=?\  \ \ \ (3)$ which cancels $(2)$
$\displaystyle\implies x^2-2xy+2y^2=(4k)^2-2(4k)+2=16k^2-8k+2=(4k-2)^2+2(4k-2)+2\  \ \ \ (4)$
$$\implies\prod_{k=1}^n\dfrac{(2k+1)^4+\dfrac14}{(2k)^4+\dfrac14}=\prod_{k=1}^n\dfrac{(4k+2)^2+2(4k+2)+2}{(4(k-1)+2)^2+2(4(k-1)+2)+2}=\prod_{k=1}^n\dfrac{f(k)}{f(k-1)}$$  where $f(m)=(4m+2)^2+2(4m+2)+2$
$$\implies\prod_{k=1}^n\dfrac{(2k+1)^4+\dfrac14}{(2k)^4+\dfrac14}=\dfrac{f(n)}{f(0)}$$
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