Question

Which one of the following is the closed form for the generating function of the sequence \( \{ a_n \}_{n \geq 0} \) defined below?
\[ a_n = \begin{cases} n+1, & \text{if } n \text{ is odd} \\ 1, & \text{otherwise} \end{cases} \]

• A. \( \dfrac{x(1 + x^2)}{(1 - x^2)^2} + \dfrac{1}{1 - x} \)
• B. \( \dfrac{x(3 - x^2)}{(1 - x^2)^2} + \dfrac{1}{1 - x} \)
• C. \( \dfrac{2x}{(1 - x^2)^2} + \dfrac{1}{1 - x} \)
• D. \( \dfrac{x}{(1 - x^2)^2} + \dfrac{1}{1 - x} \)

Hint:

When dealing with generating functions, you can split the function into parts based on conditions (like odd or even indices) and then sum them together. This approach works well for functions with alternating patterns.

Answer 1

The Correct Option is A

To find the generating function for the sequence \( a_n \), we must write the generating function for the sequence based on the given conditions. We have two conditions in the sequence: 1. If \( n \) is odd, then \( a_n = n + 1 \). 2. If \( n \) is even, then \( a_n = 1 \). Now, let's break this down by considering the even and odd cases separately: Odd-indexed terms: For odd \( n \), the term \( a_n = n + 1 \). This suggests a sequence where the value at odd positions increases linearly with \( n \). The generating function for the odd-indexed terms can be written as: \[ G_{\text{odd}}(x) = \sum_{n \text{ odd}} (n + 1) x^n \] This sum can be written as a combination of the sum for the odd powers of \( x \) and a correction term: \[ G_{\text{odd}}(x) = x + 3x^3 + 5x^5 + 7x^7 + \dots = \frac{x(1 + x^2)}{(1 - x^2)^2} \] This is derived from the sum of a geometric series where each term involves the increasing odd numbers. Even-indexed terms: For even \( n \), the term \( a_n = 1 \). The generating function for the even-indexed terms is simply: \[ G_{\text{even}}(x) = 1 + x^2 + x^4 + x^6 + \dots = \frac{1}{1 - x^2} \] Total Generating Function: Finally, to get the full generating function, we combine the generating functions for the odd and even indexed terms: \[ G(x) = G_{\text{odd}}(x) + G_{\text{even}}(x) = \frac{x(1 + x^2)}{(1 - x^2)^2} + \frac{1}{1 - x} \] This expression corresponds to Option (A). Therefore, the correct answer is (A).
Final Answer: (A)