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
(a) 4
(b) 5
(c) 6
(d) None of these

( b(1)=3 )
( b(2)=b(1+1)=b(1) f(1)=3 cdot 3=9 )
 ( b(x)=f(2+1)=f(2) b(1)=9.3=27 )

(So)
( begin{aligned} _{} b(x) operatorname{canh} 3^{x} &=3+9+27+81 =sum_{x=1}^{n} 3^{x}=120 &=b(1)+b(2)+b(3)+b(4) n=4 end{aligned} )