Question

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

• A. 4
• B. 5
• C. 6
• D. None of these

Hint:

Cauchy-type functional equation \(f(x+y)=f(x)f(y)\) gives exponential function \(f(x)=a^x\).

Answer 1

The Correct Option is A

Concept: The functional equation \(f(x+y) = f(x)f(y)\) with \(f(1) = 3\) implies \(f(n) = 3^n\).

Step 1: Identify \(f(x)\). For natural numbers, \(f(2) = f(1+1) = f(1)f(1) = 3^2\), etc. By induction, \(f(n) = 3^n\).

Step 2: Sum of the series. \[ \sum_{x=1}^{n} f(x) = 3 + 3^2 + \cdots + 3^n = \frac{3(3^n - 1)}{3 - 1} = \frac{3(3^n - 1)}{2} \]

Step 3: Set equal to 120. \[ \frac{3(3^n - 1)}{2} = 120 \Rightarrow 3(3^n - 1) = 240 \Rightarrow 3^n - 1 = 80 \Rightarrow 3^n = 81 \Rightarrow n = 4 \]