A) 4
B) 5
C) 6
D) None of these
Correct Answer: A
Solution :
Since \[f(x+y)=f(x)f(y)\] for all \[x,\ y\in N\], therefore for any \[x\in N\] \[f(x)=f(x-1+1)=f(x-1)f(1)\] \[=f(x-2){{[f(1)]}^{2}}=.......={{[f(1)]}^{x}}\] \[\Rightarrow \]\[f(x)={{3}^{x}}\], \[(\because \ f(1)=3)\] Now \[\sum\limits_{x=1}^{n}{f(x)=120}\]\[\Rightarrow \]\[\sum\limits_{x=1}^{n}{{{3}^{x}}=120}\] \[\Rightarrow \]\[\frac{3({{3}^{n}}-1)}{(3-1)}=120\]\[\Rightarrow \]\[n=4\].
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