If f is a function satisfying $f(x+y)=f(x)f(y)$ for all $x,y\in \mathbb{N}$ such that $f(1)=3$ and \(\sum_{x=1}^{n}\) f(x) = 120 , then the value of n is:

Concept:

​Geometric Progression (G.P.): A sequence where each term is obtained by multiplying the previous term by a constant called the common ratio.

Sum of n terms (S_n) of G.P. is given by: \( S_n = \frac{a(r^n - 1)}{r - 1} \)

• a: First term of the G.P.
• r: Common ratio of the G.P.
• n: Number of terms in the series

Calculation:

Given, f(1) = 3

Using the property: \( f(x + y) = f(x) \times f(y) \)

⇒ f(2) = f(1 + 1) = f(1) × f(1) = 3 × 3 = 9

⇒ f(3) = f(1 + 2) = f(1) × f(2) = 3 × 9 = 27

⇒ f(4) = f(1 + 3) = f(1) × f(3) = 3 × 27 = 81

So, \( f(1), f(2), f(3), \dots \) = 3, 9, 27, 81, … forms a G.P. with a = 3 and r = 3

Given, \( \sum_{x=1}^{n} f(x) = 120 \)

⇒ \( S_n = \frac{3(3^n - 1)}{3 - 1} \)

⇒ \( 120 = \frac{3(3^n - 1)}{2} \)

⇒ \( 120 = \frac{3}{2}(3^n - 1) \)

⇒ \( 240 = 3(3^n - 1) \)

⇒ \( 80 = 3^n - 1 \)

⇒ \( 3^n = 81 \)

⇒ \( 3^n = 3^4 \Rightarrow n = 4 \)

∴ The value of n is 4.