The coefficient of t⁴ in the expansion of ((1-t⁶)/(1-t))³ is (1) 12 (2) 15 (3) 10 (4) 14

Q: Wait, why do we ignore the $t^6$ term in the first expansion? Shouldn't we include it?

That is a fantastic question! Think about the multiplication process. We are multiplying $(1 - 3t^6 + 3t^{12} - t^{18})$ by the series expansion of $(1-t)^{-3}$, which looks like $(1 + 3t + 6t^2 + 10t^3 + 15t^4 + \dots)$. If you multiply the $-3t^6$ term by any term in the second series, the lowest power you get is $t^6$. Since we only care about $t^4$, any term with a power of $t$ greater than 4 is effectively useless. We only need the constant $1$ from the first bracket.

Q: Why can't we just use the standard binomial theorem for $(1-t)^{-3}$?

The standard binomial theorem $(a+b)^n$ is for positive integer exponents. When the exponent is negative, like $-3$, we use the generalized binomial theorem (or negative binomial expansion). The formula for the coefficient of $x^r$ in $(1-x)^{-n}$ is $\binom{n+r-1}{r}$. It is a powerful tool in your JEE arsenal for handling series expansions!

Q: Is the formula $\binom{n+r-1}{r}$ always correct for these problems?

Yes, absolutely! As long as the series is convergent (which is implied in these types of problems where we look for a specific coefficient), this formula is the standard, most efficient way to find the coefficient of $x^r$ in the expansion of $(1-x)^{-n}$. Just be careful to correctly identify $n$ and $r$!

Question

The coefficient of $t^4$ in the expansion of $(\frac{1-t^6}{1-t})^3$ is

Options:

(A) 12

(B) 15

(C) 10

(D) 14

Sigma Percentile · Accepted Answer

### Analyzing the Setup

To find the coefficient of $t^4$ in the expression $(\frac{1-t^6}{1-t})^3$, we first rewrite the expression to separate the components:

$$(1-t^6)^3 	imes (1-t)^{-3}$$

This transformation allows us to treat the problem as a product of two distinct binomial structures, making the expansion manageable.

### The Filter Strategy

We expand the first part, $(1-t^6)^3$, using the binomial theorem:

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

Since we are hunting for the coefficient of $t^4$, we must consider how these terms interact with the second part of the expression, $(1-t)^{-3}$. Any term in the first expansion with a power of $t$ greater than $4$ will result in a power of $t$ greater than $4$ when multiplied by any term in the second expansion.

Therefore, we can safely ignore $-3t^6$, $3t^{12}$, and $-t^{18}$. The only term that survives our filter is the constant $1$.

### The Power of Negative Binomials

With the first part reduced to $1$, our focus shifts entirely to the second part: $(1-t)^{-3}$. We use the general formula for the coefficient of $t^r$ in the expansion of $(1-t)^{-n}$, which is given by:

$$\binom{n+r-1}{r}$$

In this specific case, we have $n=3$ and $r=4$. Substituting these values into the formula, we obtain:

$$\binom{3+4-1}{4} = \binom{6}{4}$$

### The Final Elegance

Using the symmetry property of combinations, we know that $\binom{6}{4}$ is identical to $\binom{6}{2}$. We calculate this value as follows:

$$\frac{6 	imes 5}{2 	imes 1} = 15$$

Since the constant $1$ from our first expansion multiplies with the $15t^4$ term from our second expansion, the final coefficient is simply $1 	imes 15 = 15$.

The final answer is **15**.

Animated Solution for Mathematics - Binomial Theorem: The coefficient of $t^{4}$ in the expansion of $(\frac{1 - t ^{6}}{1 - t})^{3}$ is

Select Answer:

Visualized Solution

Analyze the Expression

Expand $(1 - t^{6})^{3}$

Filter Relevant Terms

Expand $(1 - t)^{- 3}$

Formula for $(1 - x)^{- n}$

Calculate $(r n + r - 1)$

Final Conclusion

The Sigma Insight: Binomial Theorem for any Index

Analyzing the Setup

The Filter Strategy

The Power of Negative Binomials

The Final Elegance

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Animated Solution for Mathematics - Binomial Theorem: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The coefficient of t4 in the expansion of (1−t1−t6​)3 is