Geometric Series Sum Calculation

The expression inside the first logarithm is a geometric series: $2 + \frac{2}{3} + \frac{2}{9} + \dots \infty$ This series has a first term $a = 2$ and a common ratio $r = \frac{1}{3}$. The sum to infinity ($S$) is calculated as: $S = \frac{a}{1-r} = \frac{2}{1 - \frac{1}{3}} = \frac{2}{\frac{2}{3}} = 3$. Therefore, $\log_2 \left( 2 + \frac{2}{3} + \frac{2}{9} + \dots \infty \right) = \log_2(3)$.

Logarithmic Equation Simplification

The given equation is: $\log_2 (f(x)) = \left( \log_2(3) \right) \cdot \log_3 \left( 1 + \frac{f(x)}{f\left(\frac{1}{x}\right)} \right)$ Using the change of base formula for logarithms, $\log_a b = \frac{\log_c b}{\log_c a}$, we can write $\log_3 \left( 1 + \frac{f(x)}{f\left(\frac{1}{x}\right)} \right) = \frac{\log_2 \left( 1 + \frac{f(x)}{f\left(\frac{1}{x}\right)} \right)}{\log_2(3)}$. Substituting this back into the equation gives: $\log_2 (f(x)) = \log_2(3) \cdot \frac{\log_2 \left( 1 + \frac{f(x)}{f\left(\frac{1}{x}\right)} \right)}{\log_2(3)}$ $\log_2 (f(x)) = \log_2 \left( 1 + \frac{f(x)}{f\left(\frac{1}{x}\right)} \right)$ Since the logarithm function is one-to-one, we can equate the arguments: $f(x) = 1 + \frac{f(x)}{f\left(\frac{1}{x}\right)}$ Rearranging this equation (assuming $f(x) \neq 0$): $f(x) - 1 = \frac{f(x)}{f\left(\frac{1}{x}\right)}$ Dividing by $f(x)$: $1 - \frac{1}{f(x)} = \frac{1}{f\left(\frac{1}{x}\right)}$ $1 = \frac{1}{f(x)} + \frac{1}{f\left(\frac{1}{x}\right)}$

Polynomial Function Identification

We are given that $f$ is a polynomial function. Let's test a potential polynomial form that satisfies the derived functional equation $1 = \frac{1}{f(x)} + \frac{1}{f\left(\frac{1}{x}\right)}$. Consider a polynomial of the form $f(x) = 1+x^k$ for some non-negative integer $k$ (since $f$ is a polynomial). Then $f\left(\frac{1}{x}\right) = 1 + \left(\frac{1}{x}\right)^k = 1 + x^{-k}$. Substituting into the equation: $\frac{1}{f(x)} + \frac{1}{f\left(\frac{1}{x}\right)} = \frac{1}{1+x^k} + \frac{1}{1+x^{-k}} = \frac{1}{1+x^k} + \frac{1}{\frac{x^k+1}{x^k}} = \frac{1}{1+x^k} + \frac{x^k}{1+x^k} = \frac{1+x^k}{1+x^k} = 1$. This form $f(x) = 1+x^k$ satisfies the functional equation.

Now, we use the condition $f(6) = 37$: $f(6) = 1+6^k = 37$ $6^k = 37 - 1$ $6^k = 36$ $6^k = 6^2$ This implies $k=2$. So, the polynomial function is $f(x) = 1+x^2$. This is a valid polynomial function.

Final Summation Calculation

We need to calculate $\sum_{n=1}^{10} f(n)$. Substituting $f(x) = 1+x^2$: $\sum_{n=1}^{10} f(n) = \sum_{n=1}^{10} (1+n^2)$ This sum can be split into two parts: $= \sum_{n=1}^{10} 1 + \sum_{n=1}^{10} n^2$ The sum of the constant term is: $\sum_{n=1}^{10} 1 = 10 \times 1 = 10$. The sum of the first 10 squares is calculated using the formula $\sum_{n=1}^{N} n^2 = \frac{N(N+1)(2N+1)}{6}$. For $N=10$: $\sum_{n=1}^{10} n^2 = \frac{10(10+1)(2 \cdot 10 + 1)}{6} = \frac{10 \cdot 11 \cdot 21}{6} = \frac{2310}{6} = 385$. Adding the two parts together: Total sum $= 10 + 385 = 395$.