New Year Countdown - 2 !!!

$\displaystyle \sum _{n=1} ^{12} {\dfrac {n^4}{(2n-1)(2n+1)} } = \sum _{n=1} ^{12} {\dfrac {n^4}{4n^2-1} } = \sum _{n=1} ^{12} {\dfrac {n^2(n^2-\frac{1}{4})+\frac{1}{4}n^2}{4n^2-1} }$

$= \displaystyle \sum _{n=1} ^{12} {\dfrac {n^2(n^2-\frac{1}{4})+\frac{1}{4}(n^2-\frac{1}{4}) +\frac{1}{16}}{4n^2-1} }$

$\displaystyle = \sum _{n=1} ^{12} {\dfrac {\frac{1}{4}n^2(4n^2-1)+\frac{1}{16}(4n^2-1) +\frac{1}{16}}{4n^2-1} }$

$\displaystyle = \frac {1}{4} \sum _{n=1} ^{12} {n^2} + \frac {1}{16} \sum _{n=1} ^{12} {1} + \frac {1}{16} \sum _{n=1} ^{12} {\dfrac {1}{4n^2-1}}$

$\displaystyle = \frac {1}{4} \left( \frac {12(13)(25)}{6} \right) + \frac {1}{16} (12) + \frac {1}{16} \sum _{n=1} ^{12} {\frac{1}{2} \left( \frac {1}{2n-1}-\frac{1}{2n+1} \right)}$

$\displaystyle = \frac {325}{4} + \frac {3}{4} + \frac {1}{32} \left( \sum _{n=1} ^{12} {\frac {1}{2n-1}} - \sum _{n=2} ^{13} {\frac {1}{2n-1}} \right)$

$\displaystyle = \frac {325}{4} + \frac {3}{4} + \frac {1}{32} \left( \frac{1}{1} - \frac{1}{25} \right) = \frac {325}{4} + \frac {3}{4} + \frac{3}{100} = \frac {4082} {25} = \frac {a}{b}$

$\Rightarrow a + b = \boxed{4107}$