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 $\sum\limits_{k = 1}^{10} {f(a + k)} = 16(2^{10} - 1),$ 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

The values of $b$ and $c$ for which the identity $f(x + 1) - f(x) = 8x + 3$ is satisfied,where $f(x) = bx^2 + cx + d$,are:

If a function $f$ satisfies $f(m+n) = f(m) + f(n)$ for all $m, n \in \mathbb{N}$ and $f(1) = 1$,then the largest natural number $\lambda$ such that $\sum_{k=1}^{2022} f(\lambda+k) \leq (2022)^2$ 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:

If $f : \mathbb{Z} \rightarrow \mathbb{Z}$ is defined by $f(x) = x^{9} - 11 x^{8} - 2 x^{7} + 22 x^{6} + x^{4} - 12 x^{3} + 11 x^{2} + x - 3, \forall x \in \mathbb{Z}$,then $f(11) = $