11. Find the value of $\tan \left(\tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{2 \times 2^{2}}+\tan ^{-1} \frac{1}{2 \times 3^{2}}+\cdots+\tan ^{-1} \frac{1}{2 \times 2009^{2}}\right)$.
(2 marks)求 $\tan \left(\tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{2 \times 2^{2}}+\tan ^{-1} \frac{1}{2 \times 3^{2}}+\cdots+\tan ^{-1} \frac{1}{2 \times 2009^{2}}\right)$ 的值。
11. $\frac{2009}{2010}$
11. Note that $\tan ^{-1} \frac{1}{2 k^{2}}=\tan ^{-1} \frac{(2 k+1)-(2 k-1)}{1+(2 k-1)(2 k+1)}=\tan ^{-1}(2 k+1)-\tan ^{-1}(2 k-1)$. Hence
$$
\begin{aligned}
\tan \left(\tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{2 \times 2^{2}}+\tan ^{-1} \frac{1}{2 \times 3^{2}}+\cdots+\tan ^{-1} \frac{1}{2 \times 2009^{2}}\right) & =\tan \left(\tan ^{-1}(2 \times 2009+1)-\tan ^{-1} 1\right) \\
& =\tan \left(\tan ^{-1} \frac{2 \times 2009+1-1}{1+(2 \times 2009+1)}\right) \\
& =\frac{2009}{2010}
\end{aligned}
$$

Remark. One could easily see a pattern and make a guess for the answer by computing $\tan \left(\tan ^{-1} \frac{1}{2}\right), \tan \left(\tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{2 \times 2^{2}}\right), \tan \left(\tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{2 \times 2^{2}}+\tan ^{-1} \frac{1}{2 \times 3^{2}}\right)$ and so on.