Find the coefficient of $x^4$ in the expansion of $(1 + x + x^2 + x^3)^{11}$.

Find the coefficient of x⁴ in the expansion of (1 + x + x² + x³)¹¹.

NCERT Exemplar · Q11

x^4

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Concept understanding — Binomial Theorem Expansion

The Problem That Started It All

Imagine you have to expand $(x + y)^2$ . That's easy: $x^2 + 2xy + y^2$ . Now try $(x + y)^3$ : $x^3 + 3x^2y + 3xy^2 + y^3$ . Still manageable.

But what about $(x + y)^{10}$ ? Or $(x + y)^{100}$ ? Multiplying it out term by term would take forever. There has to be a pattern — and there is.

The Binomial Theorem is the shortcut that tells you exactly what each term in the expansion of $(x + y)^n$ looks like, without ever having to multiply.

The Pattern You Already Know

Look at the expansions you already know:

Power	Expansion
$(x + y)^0$	$1$
$(x + y)^1$	$x + y$
$(x + y)^2$	$x^2 + 2xy + y^2$
$(x + y)^3$	$x^3 + 3x^2y + 3xy^2 + y^3$
$(x + y)^4$	$x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4$

Notice three things:

The powers of $x$ decrease from $n$ down to $0$ , while the powers of $y$ increase from $0$ up to $n$ . In every term, the exponents add to $n$ .
The coefficients — 1, 4, 6, 4, 1 for $n=4$ — follow a famous pattern called Pascal's triangle.
The number of terms is always $n+1$ .

Note

Pascal's triangle: each number is the sum of the two numbers directly above it.

       1
      1 1
     1 2 1
    1 3 3 1
   1 4 6 4 1
  1 5 10 10 5 1

The Precise Statement

n

Let's break that down.

The symbol $\sum_{k=0}^{n}$ means "add up terms for $k = 0, 1, 2, \dots, n$ ". For each $k$ , the term is:

$\binom{n}{k}$ — the coefficient (read as "n choose k")
$x^{n-k}$ — $x$ raised to the power $n-k$
$y^k$ — $y$ raised to the power $k$

So for $n=4$ , the terms are:

$k$	$\binom{4}{k}$	$x^{4-k}$	$y^k$	Term
0	$\binom{4}{0}=1$	$x^4$	$y^0$	$1 \cdot x^4$
1	$\binom{4}{1}=4$	$x^3$	$y^1$	$4x^3y$
2	$\binom{4}{2}=6$	$x^2$	$y^2$	$6x^2y^2$
3	$\binom{4}{3}=4$	$x^1$	$y^3$	$4xy^3$
4	$\binom{4}{4}=1$	$x^0$	$y^4$	$1 \cdot y^4$

Add them up: $x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4$ . Exactly what we had.

Where Do Those Coefficients Come From?

The binomial coefficient $\binom{n}{k}$ counts how many ways you can choose $k$ items from a set of $n$ items. In the expansion, it counts how many ways you can pick $k$ copies of $y$ (and therefore $n-k$ copies of $x$ ) when multiplying $(x+y)$ by itself $n$ times.

Tip

To compute $\binom{n}{k}$ quickly: start at $n$ and multiply $k$ decreasing numbers, then divide by $k!$ .

Example: $\binom{7}{3} = \frac{7 \cdot 6 \cdot 5}{3 \cdot 2 \cdot 1} = 35$ .

The General Term

The $k$ -th term (starting with $k=0$ ) in the expansion is:

\text{General term: } T_{k+1} = \binom{n}{k} x^{n-k} y^k

This is the most useful part for exams. If someone asks "find the 5th term in $(x+y)^{10}$ ", you set $k=4$ (because $T_{k+1}$ means $k=4$ gives the 5th term) and write:

T_5 = \binom{10}{4} x^{6} y^{4}

What If It's Not Just $x$ and $y$ ? …

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Rewrite the sum as a geometric series $\frac{1-x^4}{1-x}$ , expand using the binomial theorem on both factors, then extract the coefficient of $x^4$ by convolution. The coefficient is $990$ .

The expression $(1 + x + x^2 + x^3)^{11}$ looks unwieldy at first, but recognizing the inner sum as a geometric series transforms the problem completely.

Notice that $1 + x + x^2 + x^3 = \frac{1 - x^4}{1 - x}$ for $x \neq 1$ . This is the standard formula for a finite geometric series with first term $1$ , common ratio $x$ , and four terms.

So we need the coefficient of $x^4$ in:

\left(\frac{1-x^4}{1-x}\right)^{11} = \frac{(1-x^4)^{11}}{(1-x)^{11}}

This is much more tractable because we can expand each factor separately using the binomial theorem.

Step-by-step extraction

Expand the numerator $(1-x^4)^{11}$ By the binomial theorem:

(1-x^4)^{11} = \sum_{k=0}^{11} \binom{11}{k} (-x^4)^k = \sum_{k=0}^{11} \binom{11}{k} (-1)^k x^{4k}

The powers of $x$ that appear are $0, 4, 8, 12, \ldots, 44$ . For the coefficient of $x^4$ in the full expression, we only need terms up to $x^4$ , so only $k=0$ and $k=1$ matter:

(1-x^4)^{11} = 1 - 11x^4 + \text{higher powers}

Expand the denominator $(1-x)^{-11}$ Using the generalized binomial theorem (or the negative binomial series):

(1-x)^{-11} = \sum_{j=0}^{\infty} \binom{-11}{j} (-x)^j = \sum_{j=0}^{\infty} \binom{11+j-1}{j} x^j = \sum_{j=0}^{\infty} \binom{10+j}{j} x^j

The first few terms are:

1 + 11x + \binom{12}{2}x^2 + \binom{13}{3}x^3 + \binom{14}{4}x^4 + \cdots

Computing these binomial coefficients:

$\binom{12}{2} = 66$
$\binom{13}{3} = 286$
$\binom{14}{4} = 1001$ …

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Q.Find the coefficient of x4x^4x4 in the expansion of (1+x+x2+x3)11(1 + x + x^2 + x^3)^{11}(1+x+x2+x3)11.

Step-by-step extraction

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Q.Find the coefficient of $x^4$ in the expansion of $(1 + x + x^2 + x^3)^{11}$ .

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