If `f` is a function satisfying `f(x+y)=f(x)xxf(y)` for all `x ,y in N` such that `f(1)=3` and `sum_(x=1)^nf(x)=120 ,` find the value of `n` .

Explanation

It is given that
f(x+y)=f(x)f(y) for all x,y`in`N (1)
f(1)=3
Putting x=y=1 in (1) ,we get
f(2)=f(1+1)=f(1)f(1)=`3xx3=9`
Similarly,
`f(3)=f(1+1+1)=f(1+2)=f(1)f(2)=3xx9=27`
f(4)=f(1+3)=f(1)f(3)=`3xx27`=81
Thus, f(1),f(2),f(3),.....i.e., 3,9,27,... forms a G.P. with both the first term and common ratio equal to 3.
It is given that f(1)+f(2)+...+f(n)=120
`therefore120=(3(3^(n)-1))/(3-1)=3/2(3^(n)-1)`
`rArr3^(n)=81=3^(4)`
`therefore`n=4