જો વિધેય  $f(x+y)=f(x) f(y)$ for all $x, y \in N$ એવી રીતે વ્યાખ્યાયિત હોય કે જેથી, $f(1)=3$ અને $\sum\limits_{x = 1}^n {f\left( x \right) = 120,} $ તો $n$ નું મૂલ્ય શોધો. 

It is given that,

$f(x+y)=f(x) \times f(y)$ for all $x, y \in N$         .....$(1)$

$f(1)=3$

Taking $x=y=1$ in $(1)$

We obtain $f(1+1)=f(2)=f(1) f(1)=3 \times 3=9$

Similarly,

$f(1+1+1)=f(3)=f(1+2)=f(1) f(2)=3 \times 9=27$

$f(4)=f(1+4)=f(1) f(3)=3 \times 27=81$

$\therefore f(1), f(2), f(3), \ldots \ldots,$ that is $3,9,27, \ldots \ldots,$ forms a $G.P.$ with both the first term and common ratio equal to $3 .$

It is known that, $S_{n}=\frac{a\left(r^{n}-1\right)}{r-1}$

It is given that, $\sum\limits_{x = 1}^n {f\left( x \right) = 120,} $

$\therefore 120=\frac{3\left(3^{n}-1\right)}{3-1}$

$\Rightarrow 120=\frac{3}{2}\left(3^{n}-1\right)$

$\Rightarrow 3^{n}-1=80$

$\Rightarrow 3^{n}=81=3^{4}$

$\therefore n=4$

Thus, the value of $n$ is $4$