If f is a function satisfying f(x+y)=f(x) f(y | vedclass

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(A) It is given that $f(x+y)=f(x) f(y)$ for all $x, y \in N$ .....$(1)$.Given $f(1)=3$.Taking $x=y=1$ in $(1)$,we get $f(2)=f(1) f(1)=3 	imes 3=9$.Si

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$4$

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(A) It is given that $f(x+y)=f(x) f(y)$ for all $x, y \in N$ .....$(1)$.Given $f(1)=3$.Taking $x=y=1$ in $(1)$,we get $f(2)=f(1) f(1)=3 	imes 3=9$.Similarly,$f(3)=f(1+2)=f(1) f(2)=3 	imes 9=27$.$f(4)=f(1+3)=f(1) f(3)=3 	imes 27=81$.Thus,$f(1), f(2), f(3), \ldots$ forms a geometric progression $(G.P.)$ with first term $a=3$ and common ratio $r=3$.The sum of the first $n$ terms of a $G.P.$ is given by $S_n = \frac{a(r^n-1)}{r-1}$.Given $\sum_{x=1}^n f(x) = 120$,we have $120 = \frac{3(3^n-1)}{3-1}$.$120 = \frac{3}{2}(3^n-1)$.$80 = 3^n-1$.$3^n = 81 = 3^4$.Therefore,$n=4$.

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 $\sum\limits_{x = 1}^n {f(x) = 120}$,find the value of $n$.

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If $f(x)$ is a function satisfying $f(x + y) = f(x)f(y)$ for all $x, y \in N$ such that $f(1) = 3$ and $\sum_{x=1}^n f(x) = 120$,then the value of $n$ is

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