If a function $f(x)$ satisfies $f(x + y) = f(x) f(y)$ for all $x, y \in N$ such that $f(1) = 3$ and $\sum_{x=1}^n f(x) = 120$,then what is the value of $n$?

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:

If $f : R \to R$ is such that $f(x + y) = f(x) + f(y)$ for all $x, y \in R$,$f(1) = 7$ and $\sum_{r=1}^{n} f(r) = 14112$,then $n$ is equal to:

Suppose $f$ is a function satisfying $f(x + y) = f(x) + f(y)$ for all $x, y \in \mathbb{N}$ and $f(1) = \frac{1}{5}$. If $\sum_{n=1}^m \frac{f(n)}{n(n+1)(n+2)} = \frac{1}{12}$,then $m$ is equal to $...............$.

If $f: R \rightarrow R$ is defined as $f(x+y)=f(x)+f(y), \forall x, y \in R$ and $f(1)=10$, then, $\sum_{r=1}^n(f(r))^2=$

The function $f(x) = x - [x]$,where $[x]$ denotes the greatest integer function,is: