If a function $f$ satisfies $f(x+1)+f(x-1)=\sqrt{2} f(x)$,then $f(x+2)+f(x-2)=$

Suppose that a function $f: R \rightarrow R$ satisfies $f(x+y)=f(x) f(y)$ for all $x, y \in R$ and $f(1)=3$. If $\sum_{i=1}^{n} f(i)=363$,then $n$ is equal to

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:

Let $f: \mathbb{R} - \{0\} \rightarrow (-\infty, 1)$ be a function satisfying $f(x)f(\frac{1}{x}) = f(x) + f(\frac{1}{x})$. If $f(x)$ is a polynomial of degree $2$ and $f(K) = -2K$,then the sum of squares of all possible values of $K$ is:

The number of functions $f : \{1, 2, 3, 4\} \rightarrow \{ a \in \mathbb{Z} : |a| \leq 8 \}$ satisfying $f(n) + \frac{1}{n} f(n+1) = 1$ for all $n \in \{1, 2, 3\}$ is

$f: R \rightarrow R$ is defined by $f(x+y)=f(x)+12y, \forall x, y \in R$. If $f(1)=6$,then $\sum_{r=1}^n f(r)=$