Question:

Step 1: Identify the telescoping product identity. We recognize that the product can be simplified using the algebraic identity: $$(x^m - 1)(x^m + 1) = x^{2m} - 1$$ This suggests that a telescoping product will emerge when we multiply terms of the form $(x^{2^j} + 1)$. Step 2: Rewrite the product in terms of powers of 2. Let $m = \log_2 n$, so $n = 2^m$. The given product becomes: $$\prod_{k=1}^{m}\left(x^{\frac{2^m}{2^k}}+1\right) = \prod_{k=1}^{m}\left(x^{2^{m-k}}+1\right)$$ Step 3: Change the index of summation. Let $j = m - k$. As $k$ ranges from 1 to $m$, the index $j$ ranges from $m-1$ to 0: $$\prod_{k=1}^{m}\left(x^{2^{m-k}}+1\right) = \prod_{j=0}^{m-1}\left(x^{2^j}+1\right)$$ Step 4: Apply the telescoping product formula. We use the fundamental identity: $$(x-1)\prod_{j=0}^{m-1}\left(x^{2^j}+1\right) = x^{2^m}-1$$ Solving for the product: $$\prod_{j=0}^{m-1}\left(x^{2^j}+1\right) = \frac{x^{2^m}-1}{x-1}$$ Since $n = 2^m$, we have $2^m = n$: $$\prod_{j=0}^{m-1}\left(x^{2^j}+1\right) = \frac{x^n - 1}{x-1}$$ Step 5: Substitute $n = 2^{92}$. For $n = 2^{92}$, we have $m = 92$, so: $$\prod_{k=1}^{92}\left(x^{\frac{2^{92}}{2^k}}+1\right) = \frac{x^{2^{92}}-1}{x-1}$$ Step 6: Match with the given form. The problem states that the product equals $\dfrac{x^A - B}{x - C}$. Comparing $\dfrac{x^{2^{92}}-1}{x-1}$ with $\dfrac{x^A - B}{x - C}$: $$A = 2^{92}, \quad B = 1, \quad C = 1$$ Step 7: Calculate the final answer. $$B + C + \log_2 A = 1 + 1 + \log_2(2^{92}) = 1 + 1 + 92 = 94$$ **Final Answer: 94** The answer is **Option 3: 94**