Question

For every real number \(x \neq -1\), let \(f(x) = \frac{x}{x+1}\). Write \(f_1(x) = f(x)\) and for \(n \geq 2\), \(f_n(x) = f(f_{n-1}(x))\). Then \(f_1(-2), f_2(-2), \ldots, f_n(-2)\) must be:

• A. \(\frac{2n}{3 \cdot 1 \cdot 5 \cdot \ldots \cdot (2n-1)}\)
• B. \(1\)
• C. \(\frac{1}{2} \binom{2n}{n}\)
• D. \(\frac{2n}{n}\)

Hint:

When applying a recursive function like this, calculate the first few terms to recognize a pattern, then generalize using the discovered pattern.

Answer 1

The Correct Option is A

Step 1: The function \( f(x) = \frac{x}{x+1} \) is given, and we are asked to find a general pattern for \( f_n(x) \), where \( f_n(x) = f(f_{n-1}(x)) \) for \( n \geq 2 \).

Step 2: First, calculate the first few terms:

• For \( n = 1 \), we have \( f_1(x) = f(x) = \frac{x}{x+1} \).
• For \( n = 2 \), we apply \( f \) again to \( f_1(x) \):

\[ f_2(x) = f(f_1(x)) = f\left(\frac{x}{x+1}\right) = \frac{\frac{x}{x+1}}{\frac{x}{x+1}+1} = \frac{x}{2x+1} \]

• For \( n = 3 \), apply \( f \) again to \( f_2(x) \):

\[ f_3(x) = f(f_2(x)) = f\left(\frac{x}{2x+1}\right) = \frac{\frac{x}{2x+1}}{\frac{x}{2x+1}+1} = \frac{x}{3x+1} \]

Step 3: After observing the pattern for the first few terms, it becomes clear that:

\[ f_n(x) = \frac{x}{(2n-1)x+1} \]

Step 4: Now, substitute \( x = -2 \) into the formula for each \( n \):

\[ f_n(-2) = \frac{-2}{(2n-1)(-2)+1} \]

• For example: \[ f_1(-2) = \frac{-2}{-2+1} = 2, \quad f_2(-2) = \frac{-2}{-4+1} = \frac{2}{3}, \quad f_3(-2) = \frac{-2}{-6+1} = \frac{2}{5} \quad \text{So on...} \]

Step 5: From the general pattern \( f_n(-2) = \frac{2}{(2n-1)} \), it is clear that the product follows the form:

\[ \frac{2n}{3 \cdot 1 \cdot 5 \cdots (2n-1)}. \]

Answer 2

Step 1: The function \( f(x) = \frac{x}{x+1} \) is given, and we are asked to find a general pattern for \( f_n(x) \), where \( f_n(x) = f(f_{n-1}(x)) \) for \( n \geq 2 \).

Step 2: First, calculate the first few terms:

• For \( n = 1 \), we have \( f_1(x) = f(x) = \frac{x}{x+1} \).
• For \( n = 2 \), we apply \( f \) again to \( f_1(x) \):

\[ f_2(x) = f(f_1(x)) = f\left(\frac{x}{x+1}\right) = \frac{\frac{x}{x+1}}{\frac{x}{x+1}+1} = \frac{x}{2x+1} \]

• For \( n = 3 \), apply \( f \) again to \( f_2(x) \):

\[ f_3(x) = f(f_2(x)) = f\left(\frac{x}{2x+1}\right) = \frac{\frac{x}{2x+1}}{\frac{x}{2x+1}+1} = \frac{x}{3x+1} \]

Step 3: After observing the pattern for the first few terms, it becomes clear that:

\[ f_n(x) = \frac{x}{(2n-1)x+1} \]

Step 4: Now, substitute \( x = -2 \) into the formula for each \( n \):

\[ f_n(-2) = \frac{-2}{(2n-1)(-2)+1} \]

• For example: \[ f_1(-2) = \frac{-2}{-2+1} = 2, \quad f_2(-2) = \frac{-2}{-4+1} = \frac{2}{3}, \quad f_3(-2) = \frac{-2}{-6+1} = \frac{2}{5} \quad \text{So on...} \]

Step 5: From the general pattern \( f_n(-2) = \frac{2}{(2n-1)} \), it is clear that the product follows the form:

\[ \frac{2n}{3 \cdot 1 \cdot 5 \cdots (2n-1)}. \]