<p>The value of the expression
\[\tan\!\left(\tan^{-1}\!\left(\frac{1}{2}\right)+\tan^{-1}\!\left(\frac{2}{9}\right)+\tan^{-1}\!\left(\frac{1}{8}\right)+\tan^{-1}\!\left(\frac{2}{25}\right)+\tan^{-1}\!\left(\frac{1}{18}\right)+\cdots\cdots\infty\right)\]
is:</p>
Step-by-Step Solution
Key Concept: Recognize that each term can be decomposed using the telescoping identity tan⁻¹(a) - tan⁻¹(b) = tan⁻¹((a-b)/(1+ab)), allowing the infinite series to collapse to a simple form.
<p><strong>Step 1: Identify the pattern in terms</strong><br/>Observe the denominators: 1/2, 2/9, 1/8, 2/25, 1/18, ... <br/>Rewrite: tan⁻¹(1/(1·2)), tan⁻¹(2/(3·3)), tan⁻¹(1/(4·2)), ... <br/>Actually: tan⁻¹(1/(1·2)), tan⁻¹(1/(2·3)) + tan⁻¹(1/(3·4)), tan⁻¹(1/(4·5)), ...</p><p><strong>Step 2: Apply telescoping identity</strong><br/>Use: tan⁻¹(1/(n(n+1))) = tan⁻¹(1/n) - tan⁻¹(1/(n+1))<br/>The series becomes:<br/>[tan⁻¹(1) - tan⁻¹(1/2)] + [tan⁻¹(1/2) - tan⁻¹(1/3)] + [tan⁻¹(1/3) - tan⁻¹(1/4)] + ...</p><p><strong>Step 3: Evaluate telescoping sum</strong><br/>All intermediate terms cancel:<br/>= tan⁻¹(1) - lim(n→∞) tan⁻¹(1/n)<br/>= π/4 - 0 = π/4</p><p><strong>Step 4: Apply outer tangent function</strong><br/>tan(π/4) = 1</p><p>∴ <strong>Answer: 1</strong></p>
Correct Answer: 1