GATE CS 2016 Set 1 Q26 — Combinatorics: Counting, Recurrence Relations, Generating Functions

The coefficient of

x^{12}

\left(x^{3}+x^{4}+x^{5}+x^{6}+\dots \right)^{3}

is ___________.

View worked solution

Answer: 10

We need to find the coefficient of $x^{12}$ in the expansion of:

\left(x^{3}+x^{4}+x^{5}+x^{6}+\dots \right)^{3}

\n\n

First, we can factor out $x^3$ from the expression inside the parentheses:

x^{3}+x^{4}+x^{5}+x^{6}+\dots = x^3 \left(1 + x + x^2 + x^3 + \dots\right)

\n\n

Using the infinite geometric series sum formula, $1 + x + x^2 + \dots = \frac{1}{1-x}$ for $|x| < 1$ :

\left(x^3 \cdot \frac{1}{1-x}\right)^3 = x^9 (1-x)^{-3}

\n\n

To find the coefficient of $x^{12}$ in $x^9 (1-x)^{-3}$ , we need to find the coefficient of $x^3$ in $(1-x)^{-3}$ .

\n\n

Using the negative binomial expansion:

(1-x)^{-n} = \sum_{r=0}^{\infty} \binom{n+r-1}{r} x^r

\n\n

For $n = 3$ and $r = 3$ , the coefficient of $x^3$ is:

\binom{3+3-1}{3} = \binom{5}{3} = \frac{5 \times 4 \times 3}{3 \times 2 \times 1} = 10

\n\n

Thus, the coefficient of $x^{12}$ is 10.

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Published 10 Oct 2026

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