Question

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

Answer 1

$\because f\left(1\right)=3$ and $f(x+y)$

$=f\left(x\right).f\left(y\right)$ for all $x,yϵN$ ..... (1)

Putting $x=1$ and $y=1$ in ..... (1)

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

$\Rightarrow f\left(2\right)=3\times 3=9$

Putting $x=1$ and $y=2$, in (1)

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

$\Rightarrow f\left(3\right)=3\times 9=27$

Putting $x=1$ and $y=3$, in (1)

$f(7+3)=f\left(1\right).f\left(3\right)$

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

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

$\therefore f\left(1\right)+f\left(2\right)+f\left(3\right)+\dots \dots +f\left(n\right)=120$

$\Rightarrow {\sum }_{n=1}^{n}f\left(x\right)=3+9+27+81+\dots \dots +$ to n terms = 120

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

$\Rightarrow 3(3^n-1)=120\times 2=240$

$\Rightarrow 3^n-1=\frac{240}{3}=80$

$\Rightarrow 3^n=80+1=81=3^4$

$\Rightarrow n=4$