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?
 A: $$g(k)=\dfrac{(2k+1)^4+\dfrac14}{(2k)^4+\dfrac14}=\dfrac{(4k+2)^4+4\cdot1^4}{(4k)^4+4\cdot1^4}$$
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 $f(m)=(4m+2)^2+2(4m+2)+2$
$$\implies\prod_{k=1}^n\dfrac{(2k+1)^4+\dfrac14}{(2k)^4+\dfrac14}=\dfrac{f(n)}{f(0)}$$
A: $\ln (16241)$. Work out the first few cases, use $\ln a + \ln b = \ln ab$ and then prove by induction.
A: Let's do bit by bit.
$$\frac{(2k+1)^4+\frac{1}{4}}{16k^4+\frac{1}{4}}$$
$$=\frac{(k+\frac{1}{2})^4+\frac{1}{64}}{k^4+\frac{1}{64}}$$
Now,
$$\ln\Big( \frac{(k+\frac{1}{2})^4+\frac{1}{64}}{k^4+\frac{1}{64}} \Big)$$
$$=\ln\Big( (k+\frac{1}{2})^4+\frac{1}{64} \Big)-\ln\Big( k^4+\frac{1}{64} \Big)$$
Now bring the summation.
$$\sum^{100}_{k=1} \ln\Big( (k+\frac{1}{2})^4+\frac{1}{64} \Big)-\ln\Big( k^4+\frac{1}{64} \Big)$$
$$=\Big[\ln\Big( (1+\frac{1}{2})^4+\frac{1}{64} \Big) - \ln\Big( 1^4+\frac{1}{64} \Big)\Big] + \Big[ \ln\Big( (2+\frac{1}{2})^4+\frac{1}{64} \Big) - \ln\Big( 2^4+\frac{1}{64} \Big) \Big] + \Big[ \ln\Big( (3+\frac{1}{2})^4+\frac{1}{64} \Big) - \ln\Big( 3^4+\frac{1}{64} \Big) \Big] + \dots + \Big[ \ln\Big( (100+\frac{1}{2})^4+\frac{1}{64} \Big) - \ln\Big( 100^4+\frac{1}{64} \Big) \Big]$$
Hope you can continue after this as it's now simplified
