Binomial Theorem — Complete Guide with Solved Examples

What Is the Binomial Theorem, and Why Does It Matter?

The binomial theorem gives us a formula to expand expressions of the form $(a + b)^n$ without multiplying the brackets out $n$ times. For small $n$ like 2 or 3, you can do it by hand. But try expanding $(x + 2)^{10}$ the long way — you’ll be at it for an hour.

The theorem handles this in seconds. More importantly, it reveals a deep structure: every term in the expansion follows a predictable pattern involving binomial coefficients, which connect algebra, combinatorics, and even probability.

In JEE Main, this topic carries consistent 2–3% weightage — roughly 2 questions per paper. In CBSE Class 11, it’s a full chapter with direct questions in both the school exam and competitive papers. The good news: with the theorem itself, Pascal’s Triangle, and the general term formula, you can solve every standard question that appears.

We’ll work through everything — from the statement to the trickiest JEE-level applications.

Key Terms and Definitions

Binomial expression: Any expression with exactly two terms, like $(a + b)$ , $(x - y)$ , or $(2x + 3)$ .

Index (or exponent): The power $n$ to which the binomial is raised. In the theorem, $n$ is a non-negative integer (for the standard version we study in Class 11).

Binomial coefficient: Written as $\binom{n}{r}$ or $^nC_r$ , this counts the number of ways to choose $r$ items from $n$ . It equals:

\binom{n}{r} = \frac{n!}{r!(n-r)!}

General term: The formula for the $(r+1)$ -th term of the expansion — the most-used formula in JEE problems.

Pascal’s Triangle: A triangular array where each entry is the sum of the two entries above it. Row $n$ gives the binomial coefficients for $(a+b)^n$ .

Middle term: The term at the exact centre of the expansion. Exists uniquely when $n$ is even; when $n$ is odd, there are two middle terms.

The Binomial Theorem — Statement and Structure

(a + b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r

Expanded out:

(a+b)^n = \binom{n}{0}a^n + \binom{n}{1}a^{n-1}b + \binom{n}{2}a^{n-2}b^2 + \cdots + \binom{n}{n}b^n

Total number of terms: $n + 1$

The General Term

This is the formula you’ll use in 80% of JEE questions on this topic.

T_{r+1} = \binom{n}{r} \cdot a^{n-r} \cdot b^r

Here $r = 0, 1, 2, \ldots, n$ .

Note: $T_{r+1}$ means the $(r+1)$ -th term. So the 1st term is $T_1$ (put $r=0$ ), the 2nd term is $T_2$ (put $r=1$ ), and so on.

The Middle Term

If $n$ is even: middle term is $T_{\frac{n}{2}+1}$
If $n$ is odd: two middle terms — $T_{\frac{n+1}{2}}$ and $T_{\frac{n+3}{2}}$

Important Properties

The binomial coefficients satisfy: $\binom{n}{r} = \binom{n}{n-r}$ (symmetry)
Sum of all binomial coefficients: $\sum_{r=0}^{n} \binom{n}{r} = 2^n$
Sum of coefficients at odd positions = Sum at even positions = $2^{n-1}$
$\binom{n}{r} + \binom{n}{r-1} = \binom{n+1}{r}$ — this is Pascal’s identity

Methods and Worked Examples

Method 1 — Direct Expansion (for small n)

When to use: $n \leq 4$ or when the full expansion is asked.

Example: Expand $(x + 2)^4$ .

T_{r+1} = \binom{4}{r} x^{4-r} \cdot 2^r, \quad r = 0, 1, 2, 3, 4

$r=0$ : $\binom{4}{0} x^4 \cdot 1 = x^4$
$r=1$ : $\binom{4}{1} x^3 \cdot 2 = 8x^3$
$r=2$ : $\binom{4}{2} x^2 \cdot 4 = 24x^2$
$r=3$ : $\binom{4}{3} x \cdot 8 = 32x$
$r=4$ : $\binom{4}{4} \cdot 16 = 16$

(x + 2)^4 = x^4 + 8x^3 + 24x^2 + 32x + 16

Method 2 — Finding a Specific Term

When to use: “Find the 4th term”, “find the term independent of $x$ ”, “find the coefficient of $x^k$ ”.

Example: Find the term independent of $x$ in $\left(x - \dfrac{1}{x^2}\right)^9$ .

Here $a = x$ , $b = -\dfrac{1}{x^2}$ , $n = 9$ .

T_{r+1} = \binom{9}{r} \cdot x^{9-r} \cdot \left(-\frac{1}{x^2}\right)^r = \binom{9}{r} \cdot (-1)^r \cdot x^{9-r} \cdot x^{-2r}

T_{r+1} = \binom{9}{r}(-1)^r \cdot x^{9-r-2r} = \binom{9}{r}(-1)^r \cdot x^{9-3r}

For term independent of $x$ : $9 - 3r = 0 \Rightarrow r = 3$ .

T_4 = \binom{9}{3}(-1)^3 = 84 \times (-1) = -84

Always simplify the power of $x$ as a single expression in $r$ before setting it equal to the required value. Trying to do this mentally leads to sign errors.

Method 3 — Finding the Numerically Greatest Term

This is a Class 11 / JEE Main topic that appears occasionally. The idea: find $r$ such that $\dfrac{T_{r+1}}{T_r} \geq 1$ .

\frac{T_{r+1}}{T_r} = \frac{n - r + 1}{r} \cdot \left|\frac{b}{a}\right|

Set this $\geq 1$ and solve for $r$ .

Solved Examples

Easy — CBSE Level

Q1. Find the 6th term of $\left(2x - \dfrac{1}{3}\right)^{10}$ .

T_6 = T_{5+1} = \binom{10}{5}(2x)^5 \left(-\frac{1}{3}\right)^5

= 252 \cdot 32x^5 \cdot \left(-\frac{1}{243}\right) = 252 \cdot 32 \cdot \left(-\frac{1}{243}\right) x^5

= -\frac{8064}{243} x^5 = -\frac{896}{27}x^5

Q2. Using the binomial theorem, find the value of $(1.02)^5$ correct to 4 decimal places.

Write $(1.02)^5 = (1 + 0.02)^5$ :

(1+0.02)^5 = 1 + 5(0.02) + 10(0.02)^2 + 10(0.02)^3 + \cdots

= 1 + 0.1 + 10(0.0004) + 10(0.000008) + \cdots

= 1 + 0.1 + 0.004 + 0.00008 + \cdots \approx 1.1041

Medium — JEE Main Level

Q3. If the coefficients of $(2r+1)$ -th and $(r+2)$ -th terms in the expansion of $(1+x)^{43}$ are equal, find $r$ .

The coefficient of the $(2r+1)$ -th term is $\binom{43}{2r}$ . The coefficient of the $(r+2)$ -th term is $\binom{43}{r+1}$ .

Setting them equal: $\binom{43}{2r} = \binom{43}{r+1}$

Either $2r = r+1 \Rightarrow r = 1$ , or $2r + (r+1) = 43 \Rightarrow 3r = 42 \Rightarrow r = 14$ .

Both values are valid. Answer: $r = 1$ or $r = 14$ .

This type — “two coefficients are equal, find $r$ ” — appeared in JEE Main 2023 and CBSE Board 2022. The key insight: $\binom{n}{a} = \binom{n}{b}$ means either $a = b$ or $a + b = n$ . Memorise this.

Hard — JEE Advanced Level

Q4. Find the sum $\displaystyle\sum_{r=0}^{n} r \cdot \binom{n}{r}$ .

Why this works: We use the identity $r \cdot \binom{n}{r} = n \cdot \binom{n-1}{r-1}$ .

\sum_{r=0}^{n} r \cdot \binom{n}{r} = n \sum_{r=1}^{n} \binom{n-1}{r-1} = n \cdot 2^{n-1}

The sum equals $n \cdot 2^{n-1}$ . This result is used in probability and combinatorics problems in JEE Advanced.

Q5. Find the coefficient of $x^5$ in the expansion of $(1+x)^{21} + (1+x)^{22} + \cdots + (1+x)^{30}$ .

This is a geometric series of binomials. Coefficient of $x^5$ in each term is $\binom{21}{5}, \binom{22}{5}, \ldots, \binom{30}{5}$ .

Sum $= \binom{21}{5} + \binom{22}{5} + \cdots + \binom{30}{5}$ .

Using the hockey stick identity: $\displaystyle\sum_{i=r}^{n} \binom{i}{r} = \binom{n+1}{r+1}$ .

= \binom{31}{6} - \binom{21}{6}

= 736281 - 54264 = 682017

Exam-Specific Tips

CBSE Class 11

CBSE typically asks: direct expansion (1 mark), general term (2 marks), middle term (2 marks), and one application problem like finding the value of an expression using binomial approximation (3 marks). The full expansion of $(a+b)^4$ or $(a+b)^5$ comes with partial marking — write all terms to ensure you don’t lose marks on the coefficient.

Always show the general term formula before substituting — it earns you a step mark.
In approximation problems (like computing $\sqrt{1.04}$ using binomial), keep terms up to $x^3$ unless told otherwise.

JEE Main

In JEE Main, binomial theorem questions are almost always of these three types: (1) find the term independent of $x$ , (2) find coefficient of a specific power, (3) sum of binomial coefficients identities. The last type is increasing in frequency from 2022 onward. Practice at least 10 each of Types 1 and 2, and 5 of Type 3.

Weightage: 4 marks per question, 1–2 questions per paper.
For Type 1 (term independent of $x$ ): the answer is often an integer like $-84$ , $252$ , or $-672$ . A non-integer answer usually means you made a sign error.

JEE Advanced

At this level, questions combine binomial theorem with combinatorics or complex numbers. Common patterns: proving binomial coefficient identities, finding the greatest term, working with the sum $\displaystyle\sum r^2 \binom{n}{r}$ .

The multinomial theorem (extension to $(a+b+c)^n$ ) occasionally appears — know the general term formula: $\dfrac{n!}{p!\,q!\,r!} a^p b^q c^r$ where $p+q+r=n$ .

Common Mistakes to Avoid

Mistake 1: Wrong numbering of terms.
Students write “4th term is $T_4$ , so $r = 4$ ”. Wrong. The $(r+1)$ -th term means $T_4$ corresponds to $r = 3$ . Always subtract 1 from the term number to get $r$ .

Mistake 2: Forgetting the negative sign in binomials like $(x - y)^n$ .
When $b = -y$ , the general term has $(-y)^r$ , which is $(-1)^r y^r$ . Students often drop $(-1)^r$ , getting the sign of alternating terms wrong.

Mistake 3: Middle term confusion when $n$ is odd.
For $(a+b)^9$ , $n = 9$ is odd, so there are TWO middle terms: $T_5$ and $T_6$ . Students often report only one.

Mistake 4: Confusing $\binom{n}{r}$ with $\binom{n+1}{r}$ in Pascal’s identity.
The identity is $\binom{n}{r} + \binom{n}{r-1} = \binom{n+1}{r}$ , not $\binom{n+1}{r+1}$ . Draw it from Pascal’s Triangle until the pattern sticks.

Mistake 5: Not simplifying the combined power of $x$ before solving.
In $\left(x^2 + \dfrac{1}{x}\right)^9$ , students write $T_{r+1} = \binom{9}{r}(x^2)^{9-r}\left(\dfrac{1}{x}\right)^r$ and then forget to combine the exponents: $2(9-r) - r = 18 - 3r$ . This is where most sign and power errors happen.

Practice Questions

Q1. Find the middle term(s) in the expansion of $\left(x + \dfrac{1}{x}\right)^{10}$ .

$n = 10$ , even. Middle term is $T_6$ (i.e., $r = 5$ ).

T_6 = \binom{10}{5} x^5 \cdot x^{-5} = 252 \cdot x^0 = 252

The middle term is $252$ .

Q2. Find the coefficient of $x^3$ in the expansion of $(1 - 2x)^7$ .

T_{r+1} = \binom{7}{r}(1)^{7-r}(-2x)^r = \binom{7}{r}(-2)^r x^r

For $x^3$ : $r = 3$ .

T_4 = \binom{7}{3}(-2)^3 = 35 \times (-8) = -280

Coefficient of $x^3$ is $-280$ .

Q3. Show that $2^{10} = 1024$ using the binomial theorem.

$(1+1)^{10} = \displaystyle\sum_{r=0}^{10}\binom{10}{r} = \binom{10}{0}+\binom{10}{1}+\cdots+\binom{10}{10}$

$= 1 + 10 + 45 + 120 + 210 + 252 + 210 + 120 + 45 + 10 + 1 = 1024$

Q4. Find the term independent of $x$ in $\left(\sqrt{x} - \dfrac{2}{x}\right)^{12}$ .

T_{r+1} = \binom{12}{r}(\sqrt{x})^{12-r}\left(-\frac{2}{x}\right)^r = \binom{12}{r}(-2)^r x^{\frac{12-r}{2} - r}

Power of $x$ : $\dfrac{12-r}{2} - r = \dfrac{12-3r}{2}$ .

Set equal to 0: $12 - 3r = 0 \Rightarrow r = 4$ .

T_5 = \binom{12}{4}(-2)^4 = 495 \times 16 = 7920

Q5. If the 3rd and 4th terms of $(a + b)^n$ are 60 and 160 respectively, and the 5th term is 240, find $a$ , $b$ , $n$ .

$T_3 = \binom{n}{2}a^{n-2}b^2 = 60$

$T_4 = \binom{n}{3}a^{n-3}b^3 = 160$

$T_5 = \binom{n}{4}a^{n-4}b^4 = 240$

Dividing $T_4/T_3$ : $\dfrac{(n-2)}{3}\cdot\dfrac{b}{a} = \dfrac{8}{3}$

Dividing $T_5/T_4$ : $\dfrac{(n-3)}{4}\cdot\dfrac{b}{a} = \dfrac{3}{2}$

From these: $\dfrac{n-2}{n-3} = \dfrac{8/3}{3/2} \cdot \dfrac{4}{3} = \dfrac{32}{27} \cdot \dfrac{1}{1}$ … (solve simultaneously to get $n = 5$ , $b/a = 2$ , and using $T_3$ : $a = 1$ , $b = 2$ .)

Answer: $n = 5$ , $a = 1$ , $b = 2$ .

Q6. Find the sum: $\binom{10}{0} + \binom{10}{1} + \binom{10}{2} + \cdots + \binom{10}{5}$ .

Total sum $= 2^{10} = 1024$ . By symmetry, the first half equals the second half, except the middle term $\binom{10}{5} = 252$ is counted once.

\text{Sum} = \frac{2^{10} + \binom{10}{5}}{2} = \frac{1024 + 252}{2} = 638

Q7. Find the numerically greatest term in the expansion of $(3 - 5x)^{15}$ when $x = \dfrac{1}{5}$ .

Here $a = 3$ , $b = -5x = -1$ (substituting $x = 1/5$ ), $n = 15$ .

\frac{|T_{r+1}|}{|T_r|} = \frac{16-r}{r} \cdot \frac{1}{3}

Set $\geq 1$ : $16 - r \geq 3r \Rightarrow r \leq 4$ .

So $r = 4$ , meaning $T_5$ is the greatest term.

T_5 = \binom{15}{4}(3)^{11}(-1)^4 = 1365 \times 177147 = 241,805,655

Q8. Prove that $\binom{n}{1} + 2\binom{n}{2} + 3\binom{n}{3} + \cdots + n\binom{n}{n} = n \cdot 2^{n-1}$ .

Differentiate $(1+x)^n = \sum_{r=0}^{n}\binom{n}{r}x^r$ with respect to $x$ :

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

Put $x = 1$ :

n \cdot 2^{n-1} = \sum_{r=1}^{n} r\binom{n}{r} = \binom{n}{1} + 2\binom{n}{2} + \cdots + n\binom{n}{n}

Proved.

Frequently Asked Questions

Q: What is the total number of terms in the expansion of $(a + b + c)^n$ ?

It’s $\dfrac{(n+1)(n+2)}{2}$ . For a trinomial, we use the multinomial theorem. The general term is $\dfrac{n!}{p!\,q!\,r!}a^p b^q c^r$ where $p + q + r = n$ .

Q: Can the binomial theorem be applied when $n$ is a fraction or negative?

Yes — but only for $|x| < 1$ , and the expansion becomes an infinite series. This is called the generalised binomial theorem and is studied in higher classes. For JEE Main, you need only the finite case where $n$ is a positive integer.

Q: What is Pascal’s Triangle and how is it related to the binomial theorem?

Pascal’s Triangle lists binomial coefficients row by row. Row $n$ gives $\binom{n}{0}, \binom{n}{1}, \ldots, \binom{n}{n}$ . Each entry is the sum of the two entries directly above it (Pascal’s identity: $\binom{n}{r} = \binom{n-1}{r-1} + \binom{n-1}{r}$ ). For $n \leq 6$ , it’s faster to read coefficients off Pascal’s Triangle than to compute them.

Q: How do I find which term has the maximum value in a binomial expansion?

Compute the ratio $\dfrac{T_{r+1}}{T_r}$ and find where it transitions from $> 1$ to $< 1$ . The term at the transition point is the greatest. If $\dfrac{T_{r+1}}{T_r} = 1$ for some $r$ , then $T_{r+1}$ and $T_{r+2}$ are both the greatest (equal maximum terms).

Q: Is there a shortcut to find the sum of coefficients of a binomial expansion?

Put $x = 1$ (and $y = 1$ if there are two variables). For example, the sum of coefficients in $(3x - 2y)^{10}$ is $(3 - 2)^{10} = 1$ . The sum of coefficients of $(2x + 1)^5$ is $(2 + 1)^5 = 243$ .

Q: What is the difference between “term” and “coefficient” in these questions?

The term includes the variable part (like $252x^4$ ), while the coefficient is just the number (252). In JEE, “find the coefficient of $x^3$ ” means give the number multiplying $x^3$ , not the full term.

Q: How many PYQs from binomial theorem appear in JEE Main each year?

Typically 1–2 questions per shift, so 2–4 across a full session. From 2020–2024, the most tested subtopics were: term independent of $x$ (appeared 8 times), greatest term (4 times), and coefficient equalities (6 times). Binomial identities via differentiation/integration appeared 3 times in 2023–24 alone — this is a rising trend.

Q: Why do we need $\binom{n}{r}$ rather than just the powers of $a$ and $b$ ?

Because $\binom{n}{r}$ counts the number of ways the term $a^{n-r}b^r$ can arise when we multiply out $n$ brackets. Each bracket contributes either an $a$ or a $b$ . Choosing which $r$ brackets give $b$ (and the rest give $a$ ) is a combinations problem — and that’s exactly $\binom{n}{r}$ . The theorem isn’t just algebra; it’s combinatorics in disguise.