If f is a function satisfying f(x+y)=f(x)f(y) for all x,y ∈N such that f(1)=3 and ∑nx=1f(x)=120, find the value of n .
It is given that
f(x+y)=f(x)×f(y)∀x,y,∈N...(1)f(1)=3
Taking x=y=1 in (1), we obtain
f(1+1)=f(2)=f(1+1)=f(1)f(1)=3×3=9
Similarly
f(3)=f(1+2)=f(1)f(2)=3×9=27
f(4)=f(1+3)=f(1)f(3)=3×27=81
∴f(1), f(2), f(3),..., 3, 9, 27,.. forms
a G.P. with both the first term
and common ratio equal to 3.
It is
known that Sn=a(rn−1)r−1
Also It is given that ∑nx=1f(x)=120
∴120=3(3n−1)3−1⇒120=32(3n−1)⇒3n−1=80⇒3n=81=34∴n=4
Thus,
the value of n is 4