Question

If f is a function satisfying $f(x+y)=f\left(x\right)f\left(y\right)$ for all $x,y\in N$ such that $f\left(1\right)=3$ and $\sum _{x=1}^{n}f\left(x\right)=120$,find the value of $n$ ?

Answer 1

$f(x+y)=f\left(x\right)f\left(y\right)$
Substituting $y=1$:
$f(x+1)=3f\left(x\right)$ $(\because f(1)=3)$
$⟹\frac{f(x+1)}{f\left(x\right)}=3$
The above shows that the ratio of two conecutive terms of $f\left(x\right)$ when $x\in N$ is constant.
Therefore, the sequence $f\left(n\right)$ for natural $n$ is a GP.
$\sum _{x=1}^{n}f\left(x\right)=f\left(1\right)+f\left(2\right)+...f\left(n\right)=f\left(1\right)(1+3+3^2+..{.3}^{n-1})$
$=\frac{f\left(1\right)(3^n-1)}{3-1}=\frac{3(3^n-1)}{2}=120$
$⟹3^n=81⟹n=4$