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

To solve this problem, we need to analyze the properties of the function \( f(x) \) which satisfies the functional equation \( f(x+y) = f(x)f(y) \) with the condition \( f(1) = 3 \).

The functional equation \( f(x+y) = f(x)f(y) \) indicates that \( f(x) \) is a multiplicative function. We also have the information that \( f(1) = 3 \), which gives us a starting point.

We can hypothesize that \( f(x) \) takes the form of an exponential function, which is often the solution in such functional equations. Let's assume:

• \(f(x) = a^x\), for some constant \( a \).

Substituting \( x = 1 \) into the hypothesis, we have:

• \(f(1) = a^1 = a = 3\).

Therefore, the function can be expressed as:

• \(f(x) = 3^x\).

We know from the problem statement that \( \sum_{x=1}^{n} f(x) = 120 \). Substituting the form of \( f(x) \), we have:

• \(\sum_{x=1}^{n} 3^x = 120\).

This is a geometric series with the first term \( a = 3 \) and common ratio \( r = 3 \). The sum of the first \( n \) terms of a geometric series is given by the formula:

• \(S_n = a \frac{r^n - 1}{r - 1}\).

Applying this formula, we get:

• \(S_n = 3 \frac{3^n - 1}{3 - 1}= \frac{3^{n+1} - 3}{2}\)
• Setting this equal to 120, we solve:
• \(\frac{3^{n+1} - 3}{2} = 120\)
• \(3^{n+1} - 3 = 240\)
• \(3^{n+1} = 243\)

Since \( 243 = 3^5 \), we find that \( n+1 = 5 \), so \( n = 4 \).

Thus, the value of \( n \) that satisfies the given conditions is 4.