Question:
Suppose that a function $f: \mathrm{R} \rightarrow \mathrm{R}$ satisfies $f(\mathrm{x}+\mathrm{y})=f(\mathrm{x}) f(\mathrm{y})$ for all $\mathrm{x}, \mathrm{y} \in \mathrm{R}$ and
$f(1)=3$. If $\sum_{\mathrm{i}=1}^{\mathrm{n}} f(\mathrm{i})=363$, then $\mathrm{n}$ is equal to
Solution:
$f(\mathrm{x}+\mathrm{y})=f(\mathrm{x}) f(\mathrm{y})$
put $x=y=1 \quad f(2)=(f(1))^{2}=3^{2}$
put $x=2, y=1 \quad f(3)=(f(1))^{3}=3^{3}$
Similarly $f(\mathrm{x})=3^{\mathrm{x}}$
$\sum_{\mathrm{i}=1}^{\mathrm{n}} f(\mathrm{i})=363 \Rightarrow \sum_{\mathrm{i}=1}^{\mathrm{n}} 3^{\mathrm{i}}=363$
$\left(3+3^{2}+\ldots+3^{\mathrm{n}}\right)=363$
$\frac{3\left(3^{n}-1\right)}{2}=363$
$3^{n}-1=242 \Rightarrow 3^{n}=243$
$\Rightarrow \mathrm{n}=5$