<p>If \(\displaystyle\sum_{r=1}^{100} \sin^{-1}\!\left(\frac{1}{\sqrt{r^2+1}\sqrt{r^2+2r+2}}\right)\) is equal to \(\tan^{-1}\!\left(\dfrac{p}{q}\right)\) where \(p\) and \(q\) are co-prime, then the value of \((p+q)\) is equal to:</p>
Step-by-Step Solution
Key Concept: Recognize that sin⁻¹(1/√[(r²+1)(r²+2r+2)]) can be decomposed as a telescoping series using the identity: sin⁻¹(a) - sin⁻¹(b) = sin⁻¹(a√(1-b²) - b√(1-a²)). Specifically, observe that the expression equals tan⁻¹(r+1) - tan⁻¹(r), creating a telescoping sum.
<p><strong>Step 1:</strong> Recognize the telescoping pattern. Note that r²+2r+2 = (r+1)²+1, so the denominator is √[(r²+1)((r+1)²+1)].</p><p><strong>Step 2:</strong> Use the identity: tan⁻¹(a) - tan⁻¹(b) = sin⁻¹(1/√[(a²+1)(b²+1)]) when a-b=1. Therefore:</p><p>sin⁻¹(1/√[(r²+1)((r+1)²+1)]) = tan⁻¹(r+1) - tan⁻¹(r)</p><p><strong>Step 3:</strong> Apply the telescoping sum:</p><p>∑(r=1 to 100) [tan⁻¹(r+1) - tan⁻¹(r)] = [tan⁻¹(2) - tan⁻¹(1)] + [tan⁻¹(3) - tan⁻¹(2)] + ... + [tan⁻¹(101) - tan⁻¹(100)]</p><p><strong>Step 4:</strong> After cancellation: = tan⁻¹(101) - tan⁻¹(1) = tan⁻¹(101) - π/4</p><p><strong>Step 5:</strong> Use tan⁻¹(a) - tan⁻¹(b) = tan⁻¹((a-b)/(1+ab)):</p><p>tan⁻¹(101) - tan⁻¹(1) = tan⁻¹((101-1)/(1+101·1)) = tan⁻¹(100/102) = tan⁻¹(50/51)</p><p><strong>Step 6:</strong> Here p = 50, q = 51, which are coprime. gcd(50,51) = 1 ✓</p><p>∴ Answer: p + q = 50 + 51 = <strong>101</strong></p>
Correct Answer: D