Question

If the function $f$ satisfies the relation $f(x+y)=f\left(x\right)f\left(y\right)$ for all $x,y\in N$. Further if $f\left(1\right)=3$ and $\sum _{x=1}^{n}f\left(x\right)=120$, then the value of $n$ is:

• A. $4$
• B. $2$
• C. $1$
• D. $3$

Answer 1

The correct option is A

$4$

Explanation for the correct option

The given summation $\sum _{x=1}^{n}f\left(x\right)=120$ can be elaborated as $f\left(1\right)+f\left(2\right)+f\left(3\right)+...+f\left(n\right)=120$

According to the given condition $f(x+y)=f\left(x\right)f\left(y\right)$ and $f\left(1\right)=3$, we can evaluate the following values,

$$\begin{gathered} \Rightarrow f\left(2\right)=f(1+1)=f\left(1\right)f\left(1\right)=3\times 3=3^2 \\ \Rightarrow f\left(3\right)=f(2+1)=f\left(2\right)f\left(1\right)=3^2\times 3=3^3 \\ \Rightarrow f\left(4\right)=f(2+2)=f\left(2\right)f\left(2\right)=3^2\times 3^2=3^4 \end{gathered}$$

and so on.

Replacing these values in $f\left(1\right)+f\left(2\right)+f\left(3\right)+...+f\left(n\right)=120$, we get $3+3^2+3^3+...+3^n=120$.

Thus, a Geometric Progression is formed where $a=3,r=3,S_n=\frac{a\left(r^n-1\right)}{r-1}$

$$\begin{gathered} \Rightarrow \frac{3(3^n-1)}{3-1}=120 \\ \Rightarrow 3^n-1=80 \\ \Rightarrow 3^n=81 \\ \Rightarrow 3^n=3^4 \end{gathered}$$

Upon comparison of LHS and RHS, $n=4$

Therefore, option (A) is the correct answer i.e. $4$