Suppose that a function $f: R \rightarrow R$ satisfies $f(x+y)$ $=f(x) f(y)$ for all $x, y \in R$ an

Question

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

Professor PC Thomas Classes · Answer

(5) : We have, a function $f: R ightarrow R$ satisfies $f(x+y)=f(x) \cdot f(y) \forall x, y \in R$.\ Consider, $f(x)=a^{x}$, where $a$ is some constant.\ Then definitely we have $f(x+y)=a^{x+y}$ $=a^{x} a^{y}=f(x) \cdot f(y)$\ Now, as $f(1)=3$\ $	herefore \quad a^{1}=3 \Rightarrow a=3$.\ So, $f(x)=3^{x}$\ Now, consider $\sum_{i=1}^{n} f(i)=363$\ $\Rightarrow \quad f(1)+f(2)+\ldots .+f(n)=363$\ $\Rightarrow \quad 3^{1}+3^{2}+\ldots .+3^{n}=363$\ $\Rightarrow 3\left(\frac{3^{n}-1}{3-1}ight)=363$\ $\left[\because 3^{1}, 3^{2}, 3^{3}, \ldots, 3^{n}ight.$ is a G.P. with $a=3$ and $r=3$ ]\ $\Rightarrow \quad \frac{3^{n}-1}{2}=121$\ $\Rightarrow \quad 3^{n}=242+1=243=3^{5} \Rightarrow n=5$