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

Step $1$: Forming a G.P.

Given, $f(x+y)=f\left(x\right)f\left(y\right)x,y\in N$,

$f\left(1\right)=3$ and ${\sum }_{x=1}^{n}f\left(x\right)=120$

$f(1+1)=f\left(1\right)f\left(1\right)$

$\Rightarrow f\left(2\right)=3\times 3$

$\Rightarrow f\left(2\right)=3^2$

$f\left(3\right)=f(2+1)$

$\Rightarrow f\left(3\right)=f\left(2\right)f\left(1\right)$

$\Rightarrow f\left(3\right)=3^2\times 3$

$\Rightarrow$ $f\left(3\right)=3^3$

$f\left(4\right)=f(2+2)$

$\Rightarrow f\left(4\right)=f\left(2\right)f\left(2\right)$

$\Rightarrow$ $f\left(4\right)=3^2\times 3^2$

$\Rightarrow$ $f\left(4\right)=3^4$

$f\left(2\right)=3^2,f\left(3\right)=3^3,f\left(4\right)=3^4$

Similarly,

$f\left(5\right)=3^5$

$f\left(6\right)=3^6$

Now series is,

$3,3^2,3^3,3^4,3^5,3^6,$ ......

It’s a G.P. whose first term is $a=3$ and common ratio

$r=\frac{3^2}{3}=3$

Step $2$. Finding the value of $n$.
Sum of $n$ terms of G.P. is,

$\because S_n=\frac{a[r^n-1]}{r-1}$

Given ${\Sigma }_{x=1}^{n}f\left(x\right)=S_n=120$ & $a=3,r=3$

Putting values,

$120=\frac{3[3^n-1]}{3-1}$

$\Rightarrow$ $120\times 2=3[3^n-1]$


$\Rightarrow$ $240=3^{n+1}-3$

$\Rightarrow$ $240+3=3^{n+1}$

$\Rightarrow$$243=3^{n+1}$

$\Rightarrow$ $3^5=3^{n+1}$

Comparing of powers,

$\Rightarrow n+1=5$

$\Rightarrow n=5-1$

$\Rightarrow n=4$

Final answer: Hence, $n=4$