Question

Let $f(x)$ be a function such that $f(x+y)=f(x) \cdot f(y)$ for all $x, y \in$ N If $f(1)=3$ and $\displaystyle\sum_{k=1}^n f(k)=3279$, then the value of $n$ is

• A. 8
• B. 6
• C. 7
• D. 9

Answer 1

The Correct Option is C

Step 1: Given functional equation

The given functional equation is: 

\[ f(x + y) = f(x)f(y), \quad \forall x, y \in \mathbb{N}. \]

It is also given that \( f(1) = 3 \). Using the functional equation, we can deduce:

\[ f(2) = f(1 + 1) = f(1)f(1) = 3^2 = 9 \] \[ f(3) = f(2 + 1) = f(2)f(1) = 3^3 = 27 \] \[ f(4) = f(3 + 1) = f(3)f(1) = 3^4 = 81 \]

Thus, we observe that \( f(k) = 3^k \) for all \( k \in \mathbb{N} \).

Step 2: Using the summation formula

The sum of the series is given as:

\[ \sum_{k=1}^{n} f(k) = 3279. \]

Substitute \( f(k) = 3^k \):

\[ \sum_{k=1}^{n} 3^k = 3279. \]

This is a geometric progression with the first term \( a = 3 \), common ratio \( r = 3 \), and \( n \) terms.

The sum of a geometric progression is:

\[ S_n = \frac{a(r^n - 1)}{r - 1}. \]

Substitute the values:

\[ 3279 = \frac{3(3^n - 1)}{3 - 1}. \]

Simplifying:

\[ 3279 = \frac{3(3^n - 1)}{2}. \] \[ 3279 \times 2 = 3(3^n - 1). \] \[ 6558 = 3(3^n - 1). \] \[ 3^n - 1 = \frac{6558}{3} = 2186. \] \[ 3^n = 2187. \] 

Step 3: Solving for \( n \)

We know that \( 3^7 = 2187 \), so \( n = 7 \).