Q109 / 170 · Manipal (MET) 2020 · Mathematics
If for all $x, y \in N$, there exists a function $f(x)$ satisfying $f(x+y)=f(x) \cdot f(y)$ such that $f(1)=3$ and $\sum_{x=1}^n f(x)=120$, then value of $n$ will be
- Correct answerA$4$
- B$5$
- C$6$
- DNone of these
Solution
Given, for $x, y \in N$
$f(x+y)=f(x) \cdot f(y)$
Then, function will be of the form
$f(x)=a^x$, where $a \in N \quad[\because a \neq 1]$
$\begin{array}{ll}\because & f(1)=3 \\ \Rightarrow & f(1)=a^1=3 \\ \Rightarrow & a=3\end{array}$
$\therefore$ Function is $f(x)=3^x$.
Now, $\sum_{x=1}^n f(x)=120$
$\Rightarrow \quad \sum_{x=1}^n 3^x=120$
$\Rightarrow 3+3^2+3^3+\ldots+3^n=120$
$\Rightarrow \quad \frac{3\left(3^{n}-1\right)}{3-1}=120$
$\begin{array}{ll}\Rightarrow & 3^n=1+\frac{120 \times 2}{3} \\ \Rightarrow & 3^n=81=3^4\end{array}$
So, $n=4$
Hence, the correct answer is option: 'A'.