Question

Suppose that function $f:\mathrm{ℝ}\to \mathrm{ℝ}$ satisfies $f(x+y)=f\left(x\right)f\left(y\right)$ for all $x,y\in \mathrm{ℝ}$ and $f\left(1\right)=3$. If $\sum _{i=1}^{n}f\left(i\right)=363$, then $n$ is equal to

• A. $1$
• B. $3$
• C. $5$
• D. None of these

Answer 1

The correct option is C

$5$

Explanation for the correct option :

Step-1 : Finding the value of $f\left(0\right)$

Given that $f(x+y)=f\left(x\right)f\left(y\right)$ for all $x,y\in \mathrm{ℝ}$. Then for $x=y=0$, we get :

$$\begin{gathered} f(0+0)=f\left(0\right)f\left(0\right) \\ \Rightarrow f\left(0\right)={\left(f\left(0\right)\right)}^2 \\ \Rightarrow {\left(f\left(0\right)\right)}^2-f\left(0\right)=0 \\ \Rightarrow f\left(0\right)\left(f\left(0\right)-1\right)=0 \end{gathered}$$

This implies either $f\left(0\right)=0$ or $f\left(0\right)=1$.

If $f\left(0\right)=0$, then for any $a\in \mathrm{ℝ}$,

$$\begin{gathered} f\left(a\right)=f(a+0) \\ =f\left(a\right)f\left(0\right) \\ =0 \end{gathered}$$

which cannot be possible as $f\left(1\right)=3$ is given.

So, we are left with the only possibility that $f\left(0\right)=1$.

Step-2 : Finding $f\left(x\right)$ when $x\in \mathrm{\mathbb{N}}$ i.e. when $x\in {\mathrm{\mathbb{Z}}}^{+}$

Let $n\in \mathrm{\mathbb{N}}$ be any number. Then we can write $n=1+1+\dots +1$($n$ times). Then, we get

$$\begin{gathered} f\left(n\right)=f\left(1+1+\dots +1\right)(n\text{times}) \\ =f\left(1\right)f\left(1\right)\dots f\left(1\right)(n\text{times}) \\ =3^n\left[\because f\left(1\right)=3\right] \end{gathered}$$

Thus $f\left(n\right)=3^n$, for all $n\in \mathrm{\mathbb{N}}$.

Step-3 : Finding the value of $n$

Given that $\sum _{i=1}^{n}f\left(i\right)=363$.

Now,

$$\begin{gathered} \sum _{i=1}^{n}f\left(i\right)=f\left(1\right)+f\left(2\right)+\dots +f\left(n\right) \\ =3+3^2+\dots +3^n\left[\because f\left(n\right)=3^n\right] \\ =\frac{3\left(3^n-1\right)}{3-1}\left[\mathbf{\because }\mathit{a}\mathbf{+}\mathit{a}\mathit{r}\mathbf{+}\mathit{a}{\mathit{r}}^{\mathbf{2}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{+}\mathit{a}{\mathit{r}}^{\mathbf{n}}\mathbf{=}\frac{\mathbf{a}\left(r^n-1\right)}{\mathbf{r}\mathbf{-}\mathbf{1}}\right] \\ =\frac{3\left(3^n-1\right)}{2} \end{gathered}$$

So, we must have the following :

$$\begin{gathered} \frac{3\left(3^n-1\right)}{2}=363 \\ \Rightarrow 3^n-1=\frac{363\times 2}{3} \\ \Rightarrow 3^n=243 \\ \Rightarrow 3^n=3^5 \\ \therefore n=5 \end{gathered}$$

Hence, option (C) is the correct answer.