<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] 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
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