Find the Coefficient of t⁴ in the Expansion of [(1 − t⁶)/(1 − t)]³

JEE Main 2019 — 9 January, Evening Shift. Previous Year Question.

Problem

Find the coefficient of $t^4$ in the expansion of $\left(\dfrac{1-t^6}{1-t}\right)^3$.

Key insight. Rewriting the expression as $(1-t^6)^3\cdot(1-t)^{-3}$ splits it into two ordinary binomial expansions — and since $t^6$ already exceeds the degree being asked for, the $(1-t^6)^3$ factor contributes nothing beyond its constant term $1$ to the coefficient of $t^4$.

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Approach

Rewrite $\dfrac{1-t^6}{1-t}$ raised to the 3rd power as $(1-t^6)^3\cdot(1-t)^{-3}$. Expand $(1-t^6)^3$ using the ordinary binomial theorem — its terms are $1, -3t^6, 3t^{12}, -t^{18}$, all of degree 0 or at least 6. Since only the coefficient of $t^4$ is needed, and every term beyond the constant $1$ already has degree $\ge6$, only that constant term can combine with something from $(1-t)^{-3}$ to produce $t^4$ — so the answer reduces to reading off the coefficient of $t^4$ in the standard expansion of $(1-t)^{-3}$.

Solution

Step 1 — Rewrite the expression

$$\left(\frac{1-t^6}{1-t}\right)^3 = (1-t^6)^3\cdot(1-t)^{-3}$$

Step 2 — Expand (1 − t⁶)³

$$(1-t^6)^3 = 1-3t^6+3t^{12}-t^{18}$$

Every term here beyond the constant $1$ has degree at least 6.

Step 3 — Identify which terms can contribute to t⁴

A term $t^k$ from $(1-t^6)^3$ combines with a term $t^j$ from $(1-t)^{-3}$ to give $t^{k+j}$. Since we want $k+j=4$ and the smallest nonzero $k$ available is $6$ (already exceeding 4), only $k=0$ (the constant term $1$) can contribute — every other term would require a negative power of $t$ from the other factor, which doesn’t exist.

Step 4 — Use the negative binomial expansion of (1 − t)⁻³

The standard expansion is:

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

The coefficient of $t^4$ in this expansion is:

$$\binom{4+2}{2} = \binom62 = \frac{6\times5}{2} = 15$$

Step 5 — Combine

Since only the constant term $1$ from $(1-t^6)^3$ contributes, the coefficient of $t^4$ in the full product is exactly the coefficient of $t^4$ in $(1-t)^{-3}$:

$$1\times15 = 15$$

Answer

Common mistakes

• Trying to expand $(1-t^6)/(1-t)$ as a finite geometric series first. Writing it as $1+t+t^2+t^3+t^4+t^5$ (valid since $1-t^6=(1-t)(1+t+\cdots+t^5)$) and then cubing that six-term expression is correct but far more laborious than recognising that only degree-4-and-below terms matter at all.
• Forgetting the negative binomial coefficient formula. The coefficient of $t^n$ in $(1-t)^{-m}$ is $\binom{n+m-1}{m-1}$ — for $m=3$, this is $\binom{n+2}{2}$, not $\binom{n}{2}$ or $\binom{n+2}{3}$, and mixing up the formula gives a wrong count.
• Including the $-3t^6$ term in the sum “just in case.” Since $t^6$ alone already exceeds $t^4$, and there’s no term with a negative power of $t$ in $(1-t)^{-3}$ to cancel it down to degree 4, this term (and every term after it) genuinely contributes nothing to the coefficient of $t^4$.

Practise next

• Find the coefficient of $t^7$ in $\left(\dfrac{1-t^6}{1-t}\right)^3$.