Question:

Step 1: Apply the arc cotangent difference identity. The identity for the difference of arc cotangents is: $$\text{arc}\cot(x) - \text{arc}\cot(y) = \text{arc}\cot\left(\frac{xy+1}{y-x}\right)$$ This identity will be used to express each term of the sum as a difference. Step 2: Express the general term as a difference of arc cotangents. Consider the difference $\text{arc}\cot(2^n) - \text{arc}\cot(2^{n+1})$. Applying the identity from Step 1: $$\text{arc}\cot(2^n) - \text{arc}\cot(2^{n+1}) = \text{arc}\cot\left(\frac{2^n \cdot 2^{n+1} + 1}{2^{n+1} - 2^n}\right)$$ $$= \text{arc}\cot\left(\frac{2^{2n+1} + 1}{2^n}\right)$$ Thus, the general term of the sum can be written as: $$\text{arc}\cot\left(\frac{1+2^{2n+1}}{2^n}\right) = \text{arc}\cot(2^n) - \text{arc}\cot(2^{n+1})$$ Step 3: Evaluate the telescoping sum. Substitute the expression from Step 2 into the sum: $$\sum_{n=0}^{10} \text{arc}\cot\left(\frac{1+2^{2n+1}}{2^n}\right) = \sum_{n=0}^{10} \left[\text{arc}\cot(2^n) - \text{arc}\cot(2^{n+1})\right]$$ This is a telescoping series: $$= \left[\text{arc}\cot(2^0) - \text{arc}\cot(2^1)\right] + \left[\text{arc}\cot(2^1) - \text{arc}\cot(2^2)\right] + \cdots + \left[\text{arc}\cot(2^{10}) - \text{arc}\cot(2^{11})\right]$$ $$= \text{arc}\cot(2^0) - \text{arc}\cot(2^{11})$$ $$= \text{arc}\cot(1) - \text{arc}\cot(2^{11})$$ Step 4: Simplify the resulting expression and identify $a$ and $b$. Apply the arc cotangent difference identity to $\text{arc}\cot(1) - \text{arc}\cot(2^{11})$: $$\text{arc}\cot(1) - \text{arc}\cot(2^{11}) = \text{arc}\cot\left(\frac{1 \cdot 2^{11} + 1}{2^{11} - 1}\right)$$ $$= \text{arc}\cot\left(\frac{2^{11} + 1}{2^{11} - 1}\right)$$ Given that the sum equals $\text{arc}\cot\frac{a}{b}$, we have: $$\text{arc}\cot\frac{a}{b} = \text{arc}\cot\left(\frac{2^{11} + 1}{2^{11} - 1}\right)$$ Therefore, $a = 2^{11} + 1$ and $b = 2^{11} - 1$. $a = 2048 + 1 = 2049$ $b = 2048 - 1 = 2047$ To confirm $a$ and $b$ are coprime, we find their greatest common divisor: $\gcd(2049, 2047) = \gcd(2049 - 2047, 2047) = \gcd(2, 2047)$. Since 2047 is an odd number, $\gcd(2, 2047) = 1$. Thus, $a$ and $b$ are coprime positive integers. Step 5: Calculate the value of $\log_2\left(\dfrac{b+a}{a-b}\right)$. Substitute the expressions for $a$ and $b$: $$\frac{b+a}{a-b} = \frac{(2^{11} - 1) + (2^{11} + 1)}{(2^{11} + 1) - (2^{11} - 1)}$$ $$= \frac{2 \cdot 2^{11}}{2}$$ $$= 2^{11}$$ Now, compute the logarithm: $$\log_2\left(\frac{b+a}{a-b}\right) = \log_2(2^{11})$$ $$= 11$$