Who Wants to Sum Tan?

$tan^{-1} \frac{2}{4k^{2}}$

$=tan^{-1} \frac{2}{ 1 + 4k^{2} - 1}$

$=tan^{-1} \frac{2k + 1 - (2k - 1)}{ 1 + (2k - 1)(2k + 1)}$

$= tan^{-1}(2k + 1) - tan^{-1}(2k - 1)$

adding

$= tan^{-1}29 - tan^{-1}1$

$= tan^{-1} \frac{14}{15}$

Using the subtraction formula for tangent: $\tan {(a-b)} = \frac{\tan a - \tan b}{1 + \tan a \tan b}$ , we yield:

$\tan^{-1} u - \tan^{-1} v = \tan^{-1} \frac{u \ - \ v}{1 \ +\ uv}$ .

Then notice that: $\frac{1}{2k^2} = \frac{(2k+1) - (2k-1)}{1 + (2k+1)(2k-1)}$ . Therefore, the sum equals: $\sum_{k=1}^{14} \tan^{-1} \frac{(2k+1) - (2k-1)}{1 + (2k+1)(2k-1)} \\ = \sum_{k=1}^{14} \tan^{-1} (2k+1) - \tan^{-1} (2k-1).$

From here, we use telescoping method to get the above sum equals:

$\tan^{-1} (2(14) + 1) - \tan^{-1} (1) = \tan^{-1} \frac{(2(14)+1) - 1} {1 + (2(14)+1)} =\frac{14}{15}$

Thus, $a + b = 14 +15 = \boxed{29}$ .

GE=arctan(2k+1)-arctan(2k-1) On substitution and simplifying the telescoping series we get arctan(14/15) So 29 is the answer