Let $\sum\limits_{k = 1}^{10} {f(a + k)} = 16(2^{10} - 1),$ where the function $f$ satisfies $f(x + y) = f(x)f(y)$ for all natural numbers $x, y$ and $f(1) = 2.$ Then the natural number $a$ is

Let $f: N \rightarrow R$ be such that $f(1)=1$ and $f(1)+2 f(2)+3 f(3)+\ldots+n f(n)=n(n+1) f(n)$ for all $n \in N, n \geq 2,$ where $N$ is the set of natural numbers and $R$ is the set of real numbers. Then, the value of $f(500)$ is

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

Let $f: R \rightarrow R$ be a differentiable function that satisfies the relation $f(x + y) = f(x) + f(y) - 1$ for all $x, y \in R$. If $f'(0) = 2$,then $|f(-2)|$ is equal to:

If $f(x) = x^{2} - 3x + 4$ and $f(x) = f(2x + 1)$,then $x =$

Suppose $f : R \rightarrow (0, \infty)$ be a differentiable function such that $5f(x + y) = f(x) \cdot f(y), \forall x, y \in R$. If $f(3) = 320$,then $\sum_{n=0}^5 f(n)$ is equal to: