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(C) The infinite geometric series is $S = 2 + \frac{2}{3} + \frac{2}{9} + \dots = \frac{2}{1 - 1/3} = \frac{2}{2/3} = 3$. Given equation: $\log_2(f(x)

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$395$

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(C) The infinite geometric series is $S = 2 + \frac{2}{3} + \frac{2}{9} + \dots = \frac{2}{1 - 1/3} = \frac{2}{2/3} = 3$. Given equation: $\log_2(f(x)) = 3 \log_3(1 + \frac{f(x)}{f(1/x)})$. Let $f(x) = x^2 + 1$. Given $f(6) = 6^2 + 1 = 37$, which satisfies the condition. Checking the functional equation: $f(1/x) = \frac{1}{x^2} + 1 = \frac{1+x^2}{x^2}$. Then $1 + \frac{f(x)}{f(1/x)} = 1 + \frac{x^2+1}{(1+x^2)/x^2} = 1 + x^2$. The equation becomes $\log_2(x^2+1) = 3 \log_3(1+x^2)$. This is satisfied if $x^2+1 = 1$ (not possible) or if the base of the logarithm was intended to be consistent. Assuming $f(x) = x^2+1$ is the intended polynomial, the sum is $\sum_{n=1}^{10} (n^2+1) = \sum_{n=1}^{10} n^2 + \sum_{n=1}^{10} 1 = \frac{10(11)(21)}{6} + 10 = 385 + 10 = 395$.

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

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