Trigonometry & Inverse Trigonometry
Inverse Trigonometric Functions - Summation
Grade 11
Question:
<p>If the equation \(\sum_{n=0}^{10} \operatorname{arc}\cot\left(\frac{1+2^{2n+1}}{2^n}\right) = \operatorname{arc}\cot\frac{a}{b}\), where \(a\) and \(b\) are coprime positive integers. The value of \(\log_2\left(\frac{b+a}{a-b}\right)\), is:</p>
<p>(a) 9</p>
<p>(b) 10</p>
<p>(c) 11</p>
<p>(d) 12</p>
Step-by-Step Solution
Key Concept: Use the telescoping property of arccotangent: arccot(x) - arccot(y) = arccot((xy+1)/(y-x)). We can decompose each term as a difference of consecutive arccotangents to create a telescoping series.
<p><strong>Step 1: Decompose the general term using arccotangent identity</strong></p><p>We use the identity: arccot(x) - arccot(y) = arccot((xy+1)/(y-x)) when y > x.</p><p>For the term arccot((1+2^(2n+1))/2^n), let's check if we can write it as a difference.</p><p>Note that: arccot(2^n) - arccot(2^(n+1)) = arccot((2^n · 2^(n+1) + 1)/(2^(n+1) - 2^n))</p><p>= arccot((2^(2n+1) + 1)/(2^n(2-1))) = arccot((2^(2n+1) + 1)/2^n) ✓</p><p><strong>Step 2: Write the sum as a telescoping series</strong></p><p>∑(n=0 to 10) arccot((1+2^(2n+1))/2^n) = ∑(n=0 to 10) [arccot(2^n) - arccot(2^(n+1))]</p><p><strong>Step 3: Evaluate the telescoping sum</strong></p><p>This telescopes to: arccot(2^0) - arccot(2^11) = arccot(1) - arccot(2^11)</p><p><strong>Step 4: Use arccot difference formula to combine</strong></p><p>arccot(1) - arccot(2^11) = arccot((1·2^11 + 1)/(2^11 - 1)) = arccot((2^11 + 1)/(2^11 - 1))</p><p><strong>Step 5: Identify a and b</strong></p><p>Therefore: a/(b) = (2^11 + 1)/(2^11 - 1), so a = 2^11 + 1 = 2049 and b = 2^11 - 1 = 2047</p><p>Check gcd(2049, 2047) = gcd(2, 2047) = 1 ✓ (coprime)</p><p><strong>Step 6: Calculate the logarithm</strong></p><p>(b + a)/(a - b) = (2047 + 2049)/(2049 - 2047) = 4096/2 = 2048 = 2^11</p><p>Therefore: log₂((b+a)/(a-b)) = log₂(2^11) = 11</p><p><strong>∴ Answer: C</strong></p>
Correct Answer: C