Question

If f is a function satisfying \(f (x +y) = f(x) f(y)\text{ for all}\, x, y ∈ N\) such that f(1) = 3 and \(∑_{x = 1}^n f(x) = 120 \), find the value of n.

Answer 1

It is given that,
f (x+y) = f(x)×f (y) for all x, y∈N … (1)
f (1) = 3
Taking x = y = 1 in (1), we obtain
f (1 + 1) = f (2) = f (1) f (1) = 3 × 3 = 9
Similarly, 
f (1 + 1 + 1) = 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), …, that is 3, 9, 27, …, forms a G.P. with both the first term and common ratio equal to 3.
It is known that, \(S_n = \frac{a(rn - 1) }{ r - 1}\)
It is given that, \(∑_{x =1}^n f(x) = 120\)
∴ \(120 = \frac{3(3^n - 1)}{ 3 - 1}\)
\(⇒ 120 = \frac{3 }{2 }(3^n - 1)\)
\(⇒ 3^n - 1 = 80\)
\(⇒ 3^n = 81 = 3^4\)
∴ n = 4
Thus, the value of n is 4.