Find $\sum_{t=1}^{39} f(t)$ if $f: R \rightarrow R$ is defined as $f(x+y)=f(x)+f(y)$ for all $x, y \in R$ and $f(1)=7$.

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

Let $a, b, c \in \mathbb{R}$. If $f(x) = ax^2 + bx + c$ is such that $a + b + c = 3$ and $f(x + y) = f(x) + f(y) + xy, \forall x, y \in \mathbb{R}$,then $\sum_{n=1}^{10} f(n)$ is equal to:

If $f(1)=0$ and $f(n+1)-f(n)=5n$ for all $n \in N$, then $f(n)=$

If $f(x)$ is a polynomial function satisfying the condition $f(x) \cdot f(1/x) = f(x) + f(1/x)$ and $f(2) = 9$,then:

The number of real linear functions $f(x)$ satisfying $f(f(x))=x+f(x)$ is