We are given f (x + y) = f (x) f (y) ∀ x, y ∈ N ....(i) f (1) = 3 , $\sum\limits_{x=1}^{n}$ f (x) = 120 ∴ f (2) = f (1 + 1) = f (1) f (1) = 3·3 = $3^2$ [By using (i)] f (3) = f (2 + 1) = f (2)· f (1) = $3^2$ · 3 = $3^3$ Proceeding like above, we get f (n) = $3^n$ We have $\sum\limits_{x=1}^{n}$ f (x) = 120 ⇒ f (1) + f (2) + ...... + f (n) = 120 ⇒ 3 + $3^2$ + ...... + $3^n$ = 120 ⇒ $\frac{3(3^n-1)}{3-1}$ = 120 ⇒ $3(3^n-1)$ = 120 × 2 ⇒ $3^n - 1$ = 80 ⇒ $3^n$ = 81 ⇒ $3^n$ = $(3)^4$ ∴ n = 4