GATE CS 2017 Set 2 Q.47 (CS Q.47): Combinatorics, Counting, Recurrence Relations and Generating Functions

If the ordinary generating function of a sequence $\{a_n\}_{n=0}^\infty$ is $\frac{1+z}{(1-z)^3}$ , then $a_3 - a_0$ is equal to

Your answer

Official answer and solution

Solution

Expand the ordinary generating function $\frac{1+z}{(1-z)^3}$ using the negative binomial series to find coefficients $a_3$ and $a_0$ .

Recall the standard binomial series expansion for $(1 - z)^{-k}$ :

\frac{1}{(1-z)^k} = \sum_{n=0}^\infty \binom{n + k - 1}{n} z^n

For $k = 3$ :

\frac{1}{(1-z)^3} = \sum_{n=0}^\infty \binom{n + 2}{2} z^n = \sum_{n=0}^\infty \frac{(n+2)(n+1)}{2} z^n

Let $c_n = \frac{(n+2)(n+1)}{2}$ denote the coefficient of $z^n$ in $\frac{1}{(1-z)^3}$ .

The generating function is:

A(z) = \frac{1+z}{(1-z)^3} = \sum_{n=0}^\infty c_n z^n + z \sum_{n=0}^\infty c_n z^n = c_0 + \sum_{n=1}^\infty (c_n + c_{n-1}) z^n

Thus, the general term is $a_0 = c_0$ and $a_n = c_n + c_{n-1}$ for $n \ge 1$ .

Calculate $a_0$ and $a_3$ :

For $n = 0$ : $a_0 = c_0 = \frac{2 \times 1}{2} = 1$ .
For $n = 3$ : $c_3 = \frac{(3+2)(3+1)}{2} = \frac{5 \times 4}{2} = 10$ $c_2 = \frac{(2+2)(2+1)}{2} = \frac{4 \times 3}{2} = 6$ $a_3 = c_3 + c_2 = 10 + 6 = 16$

Compute $a_3 - a_0$ :

a_3 - a_0 = 16 - 1 = 15

Final answer

Common trap

Taking $a_3 = c_3 = 10$ and forgetting the contribution of the numerator's $+z$ term ( $z \cdot c_2 z^2 = c_2 z^3$ ).

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