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If $f$ is a function satisfying $f(x+y)=f(x) f(y)$ for all $x, y \in N$. such that $f(1)=3$ and $\sum_{x=1}^n f(x)=120$, find the value of $n$.
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$f(1)=3, f(x+y)=f(x) f(y)$
$f(2)=f(1+1)=f(1) f(1)=3.3=9$
$f(3)=f(1+2)=f(1) f(2)=3.9=27$
$f(4)=f(1+3)=f(1) f(3)=3.27=81$
Thus we have
$\sum_1^n f(x)=f(1)+f(2)+f(3)+\ldots \ldots . .+f(n)=120$
$\Rightarrow \quad 3+9+27+\ldots$. to $n$ term $=120$
or $\quad \frac{3\left(3^n-1\right)}{3-1}=120 \quad[a=3, r=3]$
$\therefore \quad \frac{3\left(3^n-1\right)}{2}=120 \Rightarrow 3^n-1=120 \times \frac{2}{3}=80$
$\quad 3^n=80+1=81=3^4 \quad \Rightarrow n=4$
$f(2)=f(1+1)=f(1) f(1)=3.3=9$
$f(3)=f(1+2)=f(1) f(2)=3.9=27$
$f(4)=f(1+3)=f(1) f(3)=3.27=81$
Thus we have
$\sum_1^n f(x)=f(1)+f(2)+f(3)+\ldots \ldots . .+f(n)=120$
$\Rightarrow \quad 3+9+27+\ldots$. to $n$ term $=120$
or $\quad \frac{3\left(3^n-1\right)}{3-1}=120 \quad[a=3, r=3]$
$\therefore \quad \frac{3\left(3^n-1\right)}{2}=120 \Rightarrow 3^n-1=120 \times \frac{2}{3}=80$
$\quad 3^n=80+1=81=3^4 \quad \Rightarrow n=4$
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