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

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 $...............$.

Let $f: R^{+} \rightarrow R^{+}$ be a function satisfying $f(x) - x = \lambda$ (constant),$\forall x \in R^{+}$ and $f(x f(y)) = f(x y) + x, \forall x, y \in R^{+}$. Then $\lim _{x \rightarrow 0} \frac{(f(x))^{\frac{1}{3}} - 1}{(f(x))^{\frac{1}{2}} - 1} =$

Let $f: R \rightarrow R$ be a twice differentiable function such that $(\sin x \cos y)(f(2x+2y) - f(2x-2y)) = (\cos x \sin y)(f(2x+2y) + f(2x-2y))$ for all $x, y \in R$. If $f'(0) = \frac{1}{2}$,then the value of $24f''\left(\frac{5\pi}{3}\right)$ is:

Let $f(x + y) = f(x) + f(y)$ for all $x, y \in R.$ Then:

If $f(x)$ is a function such that $f(x+y)=f(x)+f(y)$ and $f(1)=7$, then $\sum_{r=1}^n f(r)=$