Problem
Sum the series: \[ \tan^{-1}\left(\frac{1}{3}\right) + \tan^{-1}\left(\frac{4}{33}\right) + \cdots + \tan^{-1}\left(\frac{2^{\,n-1}}{1+2^{\,2n-1}}\right) \]
Solution
Step 1: Observe pattern
Each term can be written using identity:
\[ \tan^{-1}a – \tan^{-1}b = \tan^{-1}\left(\frac{a-b}{1+ab}\right) \]
Step 2: Apply identity
Let: \[ a = 2^k,\quad b = 2^{k-1} \]
\[ \tan^{-1}\left(\frac{2^{k-1}}{1+2^{2k-1}}\right) = \tan^{-1}(2^k) – \tan^{-1}(2^{k-1}) \]
Step 3: Write series
\[ S_n = [\tan^{-1}(2^1) – \tan^{-1}(2^0)] + [\tan^{-1}(2^2) – \tan^{-1}(2^1)] + \cdots \]
Step 4: Telescoping
All middle terms cancel:
\[ S_n = \tan^{-1}(2^n) – \tan^{-1}(1) \]
Step 5: Final result
\[ \tan^{-1}(1) = \frac{\pi}{4} \]
\[ S_n = \tan^{-1}(2^n) – \frac{\pi}{4} \]
Final Answer
\[ \boxed{\tan^{-1}(2^n) – \frac{\pi}{4}} \]
Explanation
Using telescoping series with tan⁻¹ subtraction identity simplifies the sum.