Basic Mathematics & Logarithm Question - None

Question

Let $\log_2 n$ be an integer. If $\displaystyle\prod_{k=1}^{\log_2 n}\left(x^{2^k}+1\right) = \dfrac{x^A - B}{x - C}$, where $A$, $B$ and $C$ are positive integers. Then the value of $(B + C + \log_2 A)$ for $n = 2^{92}$ is:

Accepted Answer

Step 1: Recognize the telescoping product pattern. We use the identity: (x-1)(x+1)(x²+1)(x⁴+1)...(x^(2^k)+1) = x^(2^(k+1)) - 1

Step 2: For n = 2^92, we have log₂(n) = 92. The product is ∏_{k=1}^{92}(x^(2^k)+1)

Step 3: Multiply and divide by (x-1):
∏_{k=1}^{92}(x^(2^k)+1) = [(x-1)∏_{k=1}^{92}(x^(2^k)+1)]/(x-1)

Step 4: The numerator telescopes:
(x-1)(x+1)(x²+1)(x⁴+1)...(x^(2^92)+1) = x^(2^93) - 1

Step 5: Therefore: ∏_{k=1}^{92}(x^(2^k)+1) = (x^(2^93) - 1)/(x-1)

Step 6: Comparing with (x^A - B)/(x - C):
• A = 2^93
• B = 1
• C = 1

Step 7: Calculate (B + C + log₂A):
• log₂(A) = log₂(2^93) = 93
• B + C + log₂A = 1 + 1 + 93 = 95

∴ Answer: C

Question:

Step-by-Step Solution

Master Basic Mathematics & Logarithm with Mathbee