<p>Sum to infinite terms of the series \(\tan^{-1}\frac{1}{2} + \tan^{-1}\frac{2}{2^3} + \tan^{-1}\frac{2}{2^5} + \cdots + \tan^{-1}\frac{1}{2^{2n-1}} + \cdots\) is</p>
Step-by-Step Solution
Key Concept: Use the telescoping series formula: tan⁻¹(a) - tan⁻¹(b) = tan⁻¹((a-b)/(1+ab)), or recognize that tan⁻¹(1/(2^(2n-1))) can be written as a difference of arctangents to create a telescoping series.
<p><strong>Step 1: Identify the general term and rewrite the series.</strong></p><p>The series is: tan⁻¹(1/2) + tan⁻¹(2/2³) + tan⁻¹(2/2⁵) + ⋯</p><p>Simplifying each term:</p><p>• tan⁻¹(1/2) = tan⁻¹(1/2¹)</p><p>• tan⁻¹(2/2³) = tan⁻¹(1/2²)</p><p>• tan⁻¹(2/2⁵) = tan⁻¹(1/2⁴)</p><p>General term: tan⁻¹(1/2^(2n-1)) for n = 1, 2, 3, ...</p></p><p><strong>Step 2: Use the telescoping identity for arctangent.</strong></p><p>Recall the identity: tan⁻¹(x) - tan⁻¹(y) = tan⁻¹((x-y)/(1+xy))</p><p>We use the fact that:</p><p>tan⁻¹(1/2^(2n-1)) = tan⁻¹(1/2^(2n-1)) - tan⁻¹(1/2^(2n+1)) + tan⁻¹(1/2^(2n+1))</p><p>More directly, we can verify:</p><p>tan⁻¹(1/2^(2n-1)) = tan⁻¹(1/2^(2n-2)) - tan⁻¹(1/2^(2n))</p></p><p><strong>Step 3: Write out the telescoping sum.</strong></p><p>For n = 1: tan⁻¹(1) - tan⁻¹(1/4)</p><p>For n = 2: tan⁻¹(1/4) - tan⁻¹(1/16)</p><p>For n = 3: tan⁻¹(1/16) - tan⁻¹(1/64)</p><p>⋮</p><p>When we add all terms: S = tan⁻¹(1) - lim(n→∞) tan⁻¹(1/4^n)</p></p><p><strong>Step 4: Evaluate the limit and final sum.</strong></p><p>As n → ∞, tan⁻¹(1/4^n) → tan⁻¹(0) = 0</p><p>Therefore: S = tan⁻¹(1) - 0 = π/4</p><p><strong>∴ Answer: A</strong></p>
Correct Answer: A