Question

Suppose f is a function satisfying \(f(x + y) + f(x) + f(y)\) for all \(x, y ∈N\) and \(f(1)=\frac{1}{5}\). If \(∑_{n=1}^{m} \frac{f(n)}{n(n+1)(n+2)}=\frac{1}{2}\), then m is equal to _______.

Answer 1

Correct Answer: 10

Step 1: Understanding the functional equation. The functional equation \( f(x + y) = f(x) + f(y) \) for all \( x, y \in \mathbb{N} \) suggests that \( f \) is a linear function. So, let: \[ f(x) = kx, \] where \( k \) is some constant. 

Step 2: Using the given information. We are given that \( f(1) = \frac{1}{5} \). Substituting this into the linear form: \[ f(1) = k \cdot 1 = \frac{1}{5} \quad \Rightarrow \quad k = \frac{1}{5}. \] Thus, the function becomes: \[ f(x) = \frac{x}{5}. \] 

Step 3: Substituting \( f(n) \) into the sum. We are given the equation: \[ \sum_{n=1}^{m} \frac{f(n)}{n(n + 1)(n + 2)} = \frac{1}{12}. \] Substitute \( f(n) = \frac{n}{5} \) into this sum: \[ \sum_{n=1}^{m} \frac{\frac{n}{5}}{n(n + 1)(n + 2)} = \frac{1}{12}. \] Simplifying the expression: \[ \sum_{n=1}^{m} \frac{1}{5(n + 1)(n + 2)} = \frac{1}{12}. \] Factor out \( \frac{1}{5} \): \[ \frac{1}{5} \sum_{n=1}^{m} \frac{1}{(n + 1)(n + 2)} = \frac{1}{12}. \] Multiply both sides by 5: \[ \sum_{n=1}^{m} \frac{1}{(n + 1)(n + 2)} = \frac{5}{12}. \] 

Step 4: Simplifying the sum. The sum can be simplified using partial fractions. We decompose: \[ \frac{1}{(n + 1)(n + 2)} = \frac{1}{n + 1} - \frac{1}{n + 2}. \] Thus, the sum becomes: \[ \sum_{n=1}^{m} \left( \frac{1}{n + 1} - \frac{1}{n + 2} \right). \] This is a telescoping series, and when we expand it, most terms cancel out: \[ \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \cdots + \left( \frac{1}{m + 1} - \frac{1}{m + 2} \right). \] The remaining terms are: \[ \frac{1}{2} - \frac{1}{m + 2}. \] We are given that this sum equals \( \frac{5}{12} \): \[ \frac{1}{2} - \frac{1}{m + 2} = \frac{5}{12}. \] 

Step 5: Solving for \( m \). Solve for \( m \): \[ \frac{1}{2} - \frac{1}{m + 2} = \frac{5}{12} \quad \Rightarrow \quad \frac{1}{m + 2} = \frac{1}{2} - \frac{5}{12} = \frac{1}{12}. \] Thus: \[ m + 2 = 12 \quad \Rightarrow \quad m = 10. \]