Question

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

• A. 4
• B. 5
• C. 6
• D. None of these

Answer 1

The correct option is A 4

Since $f(x+y)=f\left(x\right)f\left(y\right)$ for all $x,yϵN,$ therefore for any $xϵN,$
$f\left(x\right)=f(x-1+1)=f(x-1)f\left(1\right)$
$=f(x-2)\left[f\right(1){]}^2=...=[f\left(1\right){]}^x$
$\Rightarrow f\left(x\right)=3^x,(\because f(1)=3)$
Now ${\sum }_{x-1}^{n}f\left(x\right)=120\Rightarrow {\sum }_{x-1}^n{3}^x=120$
$\Rightarrow \frac{3(3^n-1)}{(3-1)}=120\Rightarrow n=4.$