Question

Let \(f\) be a polynomial function such that \(\log_2(f(x)) = \left(\log_2\left(2 + \frac{2}{3} + \frac{2}{9} + \dots \infty\right)\right) \cdot \log_3\left(1 + \frac{f(x)}{f(1/x)}\right)\), \(x>0\) and \(f(6) = 37\). Then \(\sum_{n=1}^{10} f(n)\) is equal to ________.

Answer 1

Correct Answer: 395

Step 1: Understanding the Concept:
We first simplify the infinite geometric series in the exponent, then use logarithmic properties to establish a functional equation for \(f(x)\). For polynomials, the identity \(f(x)f(1/x) = f(x) + f(1/x)\) often leads to the form \(f(x) = x^n + 1\).
Step 2: Key Formula or Approach:
1. Infinite G.P. sum: \(S_\infty = \frac{a}{1-r}\).
2. Property: If \(f(x)\) is a polynomial such that \(f(x) + f(1/x) = f(x)f(1/x)\), then \(f(x) = \pm x^n + 1\).
Step 3: Detailed Explanation:
Sum of G.P. \(= \frac{2}{1 - 1/3} = \frac{2}{2/3} = 3\). The equation becomes: \[ \log_2(f(x)) = \log_2(3) \cdot \log_3\left(1 + \frac{f(x)}{f(1/x)}\right) \] Using the base change formula \(\log_2 3 \cdot \log_3 Z = \log_2 Z\): \[ \log_2(f(x)) = \log_2\left(1 + \frac{f(x)}{f(1/x)}\right) \] \[ f(x) = 1 + \frac{f(x)}{f(1/x)} \implies f(x) \cdot f(1/x) = f(1/x) + f(x) \] This is the standard functional equation for polynomials, implying \(f(x) = x^n + 1\). Given \(f(6) = 37 \implies 6^n + 1 = 37 \implies 6^n = 36 \implies n = 2\). So, \(f(x) = x^2 + 1\). Now calculate the sum: \[ \sum_{n=1}^{10} (n^2 + 1) = \sum_{n=1}^{10} n^2 + \sum_{n=1}^{10} 1 \] \[ = \frac{10(11)(21)}{6} + 10 = 385 + 10 = 395 \]
Step 4: Final Answer:
The sum is 395.