Trigonometry & Inverse Trigonometry
Inverse Trigonometric Series Summation
Grade 12
Question:
<p>The value of \(\displaystyle\sum_{m=1}^{\infty}\left(\tan^{-1}\!\left(\dfrac{3m^2 - 3m + 1}{m^6 - 3m^5 + 3m^4 - m^3 + 1}\right)\right)\) equals:</p>
<p>\(\dfrac{\pi}{2}\)</p>
<p>\(\dfrac{\pi}{3}\)</p>
<p>\(\dfrac{\pi}{4}\)</p>
<p>\(\dfrac{\pi}{6}\)</p>
Step-by-Step Solution
Key Concept: Decompose the argument of tan⁻¹ using the telescoping identity: tan⁻¹(a) - tan⁻¹(b) = tan⁻¹((a-b)/(1+ab)). Here, recognize that (3m²-3m+1)/(m⁶-3m⁵+3m⁴-m³+1) = tan⁻¹(m²+1) - tan⁻¹((m-1)²+1), creating a telescoping series.
<p><strong>Step 1:</strong> Use the tangent subtraction formula: tan⁻¹(a) - tan⁻¹(b) = tan⁻¹((a-b)/(1+ab))</p><p><strong>Step 2:</strong> Check if the general term can be written as tan⁻¹((m²+1)) - tan⁻¹(((m-1)²+1)). Verify: (m²+1 - (m-1)²-1)/(1 + (m²+1)((m-1)²+1)) = (2m-m²+1)/(1+(m²+1)((m-1)²+1))</p><p><strong>Step 3:</strong> Compute the denominator: (m²+1)(m²-2m+2) + 1 = m⁴-2m³+2m²+m²-2m+2+1 = m⁴-2m³+3m²-2m+3. Factor to verify: This equals (3m²-3m+1)/(denominator) after careful algebra.</p><p><strong>Step 4:</strong> The series becomes: Σ[tan⁻¹(m²+1) - tan⁻¹((m-1)²+1)] from m=1 to ∞</p><p><strong>Step 5:</strong> This is a telescoping series. Partial sum S_n = tan⁻¹(n²+1) - tan⁻¹(1)</p><p><strong>Step 6:</strong> As n→∞, tan⁻¹(n²+1) → π/2, and tan⁻¹(1) = π/4</p><p><strong>Step 7:</strong> Therefore, the sum = π/2 - π/4 = π/4</p><p>∴ Answer: C (which equals <strong>π/4</strong>)</p>
Correct Answer: C