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: Sum GP \[ 2+\frac23+\frac29+\cdots \] \[ S=\frac{2}{1-1/3}=3 \] So equation becomes \[ \log_2 f(x)=\log_2 3\cdot \log_3\left(1+\frac{f(x)}{f(1/x)}\right) \] \[ =\log_2\left(1+\frac{f(x)}{f(1/x)}\right) \] Hence \[ f(x)=1+\frac{f(x)}{f(1/x)} \] \[ f(x)f(1/x)=f(x)+f(1/x) \] Polynomial solution: \[ f(x)=x^n+1 \] Given \[ f(6)=37 \] \[ 6^n+1=37 \] \[ 6^n=36 \] \[ n=2 \] So \[ f(x)=x^2+1 \] Step 2: Required sum \[ \sum_{n=1}^{10}f(n)=\sum_{n=1}^{10}(n^2+1) \] \[ =\sum_{n=1}^{10}n^2+10 \] \[ =\frac{10\cdot11\cdot21}{6}+10 \] \[ =385+10 \] \[ =395 \] \[ \boxed{395} \]