Sum of Terms of Polynomial Function Satisfying Logarithmic Equation

To solve the given functional equation, we first simplify the infinite geometric series in the argument of the logarithm: $2 + \frac{2}{3} + \frac{2}{9} + \dots = \frac{2}{1 - 1/3} = 3$

Substituting this back into the equation gives: $\log_2(f(x)) = \log_2(3) \cdot \log_3\left(1 + \frac{f(x)}{f(1/x)}\right) = \log_2\left(1 + \frac{f(x)}{f(1/x)}\right)$

Equating the arguments of the logarithms leads to: $f(x) = 1 + \frac{f(x)}{f(1/x)} \implies f(x) f(1/x) = f(x) + f(1/x)$

The polynomial solutions to this standard functional equation are of the form $f(x) = 1 \pm x^n$. Using the given condition $f(6) = 37$, we find $1 + 6^n = 37 \implies n = 2$, so $f(x) = x^2 + 1$.

Finally, we compute the required sum: $\sum_{n=1}^{10} f(n) = \sum_{n=1}^{10} (n^2 + 1) = \frac{10 \times 11 \times 21}{6} + 10 = 385 + 10 = 395$