The coefficient of x⁴ in the expansion of (1 + x + x² + x³)⁶ in powers of x, is

Q: Why can't we just use the multinomial theorem directly? It seems like it would work.

You are absolutely right that the multinomial theorem is a valid tool! However, in competitive exams like JEE Advanced, time is your most precious resource. Expanding $(1 + x + x^2 + x^3)^6$ using the multinomial theorem would involve calculating many terms and summing them up, which is prone to arithmetic errors. The geometric progression method is an 'elegant shortcut' that transforms a complex polynomial into a product of two simpler binomials, saving you time and mental energy.

Q: Wait, how do we expand $(1-x)^{-6}$? I only know the binomial theorem for positive integer powers.

That is a fantastic question! When the exponent is negative, we use the generalized binomial theorem. For $(1-x)^{-n}$, the expansion is $1 + nx + \frac{n(n+1)}{2!}x^2 + \dots + \binom{n+r-1}{r}x^r$. In your case, $n=6$. So, the coefficient of $x^r$ is $\binom{6+r-1}{r}$, which simplifies to $\binom{5+r}{r}$. It is a powerful tool to have in your arsenal for series expansions!

Q: Why do we ignore terms higher than $x^4$ in the first bracket?

Because we are only hunting for the coefficient of $x^4$! If you multiply a term like $x^8$ (which appears in the expansion of $(1-x^4)^6$) by any term in the second bracket, the resulting power of $x$ will be at least $x^8$. Since we only care about $x^4$, those higher-order terms are irrelevant. Focusing only on what matters is the hallmark of a smart problem solver.

Question

The coefficient of $x^4$ in the expansion of $(1 + x + x^2 + x^3)^6$ in powers of $x$, is

Sigma Percentile · Accepted Answer

### The Art of Elegant Expansion

Welcome, fellow traveler on the path to JEE mastery. Today, we are going to dismantle a problem that, at first glance, looks like a tedious exercise in brute-force algebra.

We are tasked with finding the coefficient of $x^4$ in the expansion of $(1 + x + x^2 + x^3)^6$. If you were to sit down and multiply this out by hand, you would be lost in a sea of terms for hours. But we are not here to do brute force; we are here to find the hidden structure.

### Phase 1: The Geometric Insight

Look at the expression inside the parentheses: $1 + x + x^2 + x^3$. This is a finite geometric progression with the first term $a = 1$, common ratio $r = x$, and $n = 4$ terms.

The sum of a geometric progression is given by the formula:

$$S_n = \frac{a(1-r^n)}{1-r}$$

Substituting our values, we get:

$$1 + x + x^2 + x^3 = \frac{1-x^4}{1-x}$$

Suddenly, the terrifying polynomial has been tamed into a simple fraction. Our original expression now becomes:

$$\left( \frac{1-x^4}{1-x} ight)^6 = (1-x^4)^6 (1-x)^{-6}$$

This is the turning point of the problem. We have transformed a multinomial expansion into the product of two binomials.

### Phase 2: The Binomial Dance

Now, we have two distinct parts to expand. First, consider $(1-x^4)^6$.

Using the standard binomial theorem:

$$(1-x^4)^6 = \binom{6}{0} - \binom{6}{1}x^4 + \binom{6}{2}x^8 - \dots$$

Since we only need the coefficient of $x^4$, we can stop early. The relevant terms are $1 - 6x^4$. Anything with $x^8$ or higher is irrelevant to our goal.

Next, we tackle $(1-x)^{-6}$. This is a negative binomial expansion where the general term is given by $\binom{n+r-1}{r}x^r$ with $n=6$.

For $r=0, 1, 2, 3, 4$, the terms are:

$$1, \binom{6}{1}x, \binom{7}{2}x^2, \binom{8}{3}x^3, \binom{9}{4}x^4$$

Calculating these values, we obtain:

$$1 + 6x + 21x^2 + 56x^3 + 126x^4$$

### Phase 3: The Final Selection

We are now at the finish line. We need to multiply $(1 - 6x^4)$ by $(1 + 6x + 21x^2 + 56x^3 + 126x^4)$ and extract the $x^4$ term.

There are only two ways to create an $x^4$ term:

1. Multiply the constant $1$ from the first bracket by the $126x^4$ term from the second bracket: $1 	imes 126x^4 = 126x^4$.
2. Multiply the $-6x^4$ term from the first bracket by the constant $1$ from the second bracket: $-6x^4 	imes 1 = -6x^4$.

Adding these together, we get $126x^4 - 6x^4 = 120x^4$. The final coefficient is **120**.

### Conclusion

Look at what we have achieved. By identifying the geometric progression and utilizing the negative binomial expansion, we bypassed the chaos of multinomial expansion.

This is the beauty of mathematics—it rewards those who look for patterns rather than those who simply grind through calculations. Keep this mindset, and you will find that even the most daunting JEE problems have a path to an elegant solution.

Animated Solution for Mathematics - Binomial Theorem: The coefficient of $x^{4}$ in the expansion of $(1 + x + x^{2} + x^{3})^{6}$ in powers of $x$ , is

Enter Numerical Value:

Visualized Solution

Analyze the Expression $(1 + x + x^{2} + x^{3})^{6}$

Identify the Inner Geometric Progression

Apply GP Sum Formula

Simplify the Whole Expression

Expand $(1 - x^{4})^{6}$

Expand $(1 - x)^{- 6}$

Product of the Two Expansions

Identify Terms for $x^{4}$

Calculate $(4 9)$

Final Result

The Sigma Insight: Binomial Theorem for any Index

The Art of Elegant Expansion

Phase 1

Phase 2

Phase 3

Conclusion

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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 x4 in the expansion of (1+x+x2+x3)6 in powers of x, is