GATE CS 2022 Q.36: Combinatorics, Counting, Recurrence Relations and Generating Functions

Which one of the following is the closed form for the generating function of the sequence $\{a_n\}_{n \ge 0}$ defined below?

a_n = \begin{cases} n + 1, & n \text{ is odd} \\ 1, & \text{otherwise} \end{cases}

$\frac{x(1 + x^2)}{(1 - x^2)^2} + \frac{1}{1 - x}$

$\frac{x(3 - x^2)}{(1 - x^2)^2} + \frac{1}{1 - x}$

$\frac{2x}{(1 - x^2)^2} + \frac{1}{1 - x}$

$\frac{x}{(1 - x^2)^2} + \frac{1}{1 - x}$

Official answer and solution

Solution

Decompose the sequence into a constant base sequence and an odd-indexed linear term.

The sequence is defined as:

a_n = \begin{cases} n + 1, & n \text{ is odd} \\ 1, & n \text{ is even} \end{cases}

We can decompose $a_n$ as $a_n = 1 + b_n$ , where:

b_n = \begin{cases} n, & n \text{ is odd} \\ 0, & n \text{ is even} \end{cases}

The generating function for the constant sequence $1, 1, 1, \dots$ is:

\sum_{n=0}^\infty x^n = \frac{1}{1 - x}

For $b_n$ , only the odd terms are non-zero: $\sum_{k=0}^\infty (2k + 1) x^{2k+1} = x \sum_{k=0}^\infty (2k + 1) (x^2)^k$ .

Letting $y = x^2$ :

\sum_{k=0}^\infty (2k + 1) y^k = 2 \sum_{k=0}^\infty k y^k + \sum_{k=0}^\infty y^k = 2 \frac{y}{(1 - y)^2} + \frac{1}{1 - y} = \frac{2y + (1 - y)}{(1 - y)^2} = \frac{1 + y}{(1 - y)^2} = \frac{1 + x^2}{(1 - x^2)^2}

Multiplying by $x$ gives $\frac{x(1 + x^2)}{(1 - x^2)^2}$ . Adding the constant component $\frac{1}{1-x}$ :

A(x) = \frac{x(1 + x^2)}{(1 - x^2)^2} + \frac{1}{1 - x}

This matches Option (A).

Final answer

Option (A)

Common trap

Omitting the $x$ factor in front of the odd power series.

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