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

It is given that
$f(x+y)=f\left(x\right)\times f\left(y\right)\forall x,y,\in N...\left(1\right)f\left(1\right)=3$
Taking $x=y=1$ in (1), we obtain
$f(1+1)=f\left(2\right)=f(1+1)=f\left(1\right)f\left(1\right)=3\times 3=9$

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

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

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

It is known that $S_n=\frac{a(r^n-1)}{r-1}$

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

$\therefore 120=\frac{3(3^n-1)}{3-1}\Rightarrow 120=\frac{3}{2}(3^n-1)\Rightarrow 3^n-1=80\Rightarrow 3^n=81=3^4\therefore n=4$
Thus, the value of $n$ is $4$