Concept:
Use the formula \(\tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right)\) for \(xy < 1\) to combine the terms stepwise, reducing the sum to a single inverse tangent.
Step 1: Combine the first two terms.
\[
S = \tan^{-1}\frac{1}{2} + \tan^{-1}\frac{1}{8} + \tan^{-1}\frac{1}{18} + \tan^{-1}\frac{1}{32}.
\]
First, \(\tan^{-1}\frac{1}{2} + \tan^{-1}\frac{1}{8} = \tan^{-1}\left(\frac{\frac{1}{2} + \frac{1}{8}}{1 - \frac{1}{2}\cdot\frac{1}{8}}\right) = \tan^{-1}\left(\frac{\frac{5}{8}}{\frac{15}{16}}\right) = \tan^{-1}\frac{2}{3}\).
Step 2: Combine the last two terms.
\[
\tan^{-1}\frac{1}{18} + \tan^{-1}\frac{1}{32} = \tan^{-1}\left(\frac{\frac{1}{18} + \frac{1}{32}}{1 - \frac{1}{18}\cdot\frac{1}{32}}\right) = \tan^{-1}\left(\frac{\frac{25}{288}}{\frac{575}{576}}\right) = \tan^{-1}\frac{2}{23}.
\]
Step 3: Combine the two results.
\[
S = \tan^{-1}\frac{2}{3} + \tan^{-1}\frac{2}{23} = \tan^{-1}\left(\frac{\frac{2}{3} + \frac{2}{23}}{1 - \frac{2}{3}\cdot\frac{2}{23}}\right) = \tan^{-1}\left(\frac{\frac{52}{69}}{\frac{65}{69}}\right) = \tan^{-1}\frac{4}{5}.
\]
Step 4: Write the final answer.
\[
\boxed{\tan^{-1}\left(\frac45\right)}
\]