How much of JEE Main comes from Binomial Theorem?

JEEGenie estimates about 3% of JEE Main marks come from Binomial Theorem, based on recent papers. The share varies from year to year.

Which formulas matter most in Binomial Theorem?

The Binomial Theorem sheet on this page has 23 formula entries, starting with Binomial Theorem Statement, General Term in Binomial Expansion, Finding Term Independent of x, Middle Term(s) in Binomial Expansion, Binomial Coefficient Properties. Each one comes with when to use it and the trap to avoid.

What topics does Binomial Theorem cover for JEE?

Binomial Theorem covers Binomial Expansion, General and Middle Term, Binomial Coefficients, Properties of Coefficients, Applications of Binomial Theorem.

How should I revise Binomial Theorem before the exam?

Learn each formula together with its condition of use, read the trap listed beside it, then practise questions and review every mistake. The practice link on this page opens Binomial Theorem questions on JEEGenie.

How many previous-year questions are there on Binomial Theorem?

JEEGenie has 350 previous-year questions on Binomial Theorem, and this page shows a few with worked explanations.

Is practice on Binomial Theorem free on JEEGenie?

Sign-up is free and includes a 7-day Pro trial. After the trial, free accounts keep a limited daily quota.

Binomial Theorem for JEE Main & Advanced — formulas, what is asked, and traps

JEEGenie estimates about 3% of JEE Main marks come from Binomial Theorem. This page covers 23 formula entries, each with when to use it, what JEE asks and the traps. A common slip: Forgetting that the sum runs from $r = 0$ to $n$ , not $1$ to $n$ .

Key points to know

Binomial Theorem Statement. Expands $(a + b)^{n}$ into a sum of terms involving binomial coefficients, for any positive integer $n$ .
General Term in Binomial Expansion. The $(r + 1)$ -th term in the expansion of $(a + b)^{n}$ is given by $T_{r + 1}$ .
Binomial Coefficient Notation. $(\binom{n}{r})$ represents the number of ways to choose $r$ items from $n$ , and is a key part of binomial expansion.
Pascal's Triangle. A triangular array of binomial coefficients where each entry is the sum of the two above it.

Formula sheet

Name	Formula	When to use
Binomial Theorem Statement	$(a + b)^{n} = \sum_{r = 0}^{n} (\binom{n}{r}) a^{n - r} b^{r}$	When expanding any binomial expression $(a + b)^{n}$ with a positive integer exponent.
General Term in Binomial Expansion	$T_{r + 1} = (\binom{n}{r}) a^{n - r} b^{r}$	To find any specific term in the expansion without writing all terms.
Finding Term Independent of x	For $(a x^{p} + b x^{- q})^{n}$ , $T_{r + 1} = (\binom{n}{r}) (a x^{p})^{n - r} (b x^{- q})^{r}$ , exponent of $x$ is $p (n - r) - q r$ . Set $p (n - r) - q r = 0$ .	When a problem asks for the term that does not contain $x$ (constant term) in a binomial expansion.
Middle Term(s) in Binomial Expansion	If $n$ is even: middle term $T_{n / 2 + 1}$ . If $n$ is odd: middle terms $T_{(n + 1) / 2}$ and $T_{(n + 3) / 2}$ .	When asked to find the middle term(s) in the expansion of $(a + b)^{n}$ .
Binomial Coefficient Properties	$(\binom{n}{r}) = (\binom{n}{n - r})$ ; $\sum_{r = 0}^{n} (\binom{n}{r}) = 2^{n}$ ; $\sum_{r = 0}^{n} (- 1)^{r} (\binom{n}{r}) = 0$ ; $\sum_{r = 0}^{n} {(\binom{n}{r})}^{2} = (\binom{2 n}{n})$ .	To simplify sums involving binomial coefficients, especially in combinatorial identities.
Greatest Term in Binomial Expansion	For $(1 + x)^{n}$ , $\frac{T_{r + 1}}{T_{r}} = \frac{n - r + 1}{r} ∣ x ∣$ . Solve $\frac{n - r + 1}{r} ∣ x ∣ \geq 1$ to find $r$ .	When asked to find the numerically greatest term in the expansion of $(1 + x)^{n}$ or $(a + b)^{n}$ .
Binomial Theorem for Any Index	$(1 + x)^{n} = \sum_{r = 0}^{\infty} \frac{n (n - 1) \dots (n - r + 1)}{r!} x^{r}$ for $∣ x ∣ < 1$ .	When expanding $(1 + x)^{n}$ where $n$ is not a positive integer, e.g., $\sqrt{1 + x}$ or $1 / (1 - x)$ .
Approximations Using Binomial Theorem	$(1 + x)^{n} = 1 + n x + \frac{n (n - 1)}{2} x^{2} + O (x^{3})$	When estimating values like $(1.01)^{5}$ or $(0.99)^{3}$ without a calculator.
Finding Coefficient of a Specific Power	For $(a x^{p} + b x^{q})^{n}$ , exponent of $x$ in $T_{r + 1}$ is $p (n - r) + q r$ . Set equal to $k$ and solve for $r$ .	When asked for the coefficient of $x^{m}$ in a binomial expansion.
Binomial Coefficient Notation	$(\binom{n}{r}) = \frac{n!}{r! (n - r)!}$	Whenever binomial coefficients appear in expansions or combinatorial problems.
Sum of Coefficients in Binomial Expansion	$\sum_{r = 0}^{n} (\binom{n}{r}) = 2^{n}$	To quickly find the sum of all coefficients in any binomial expansion.
Sum of Coefficients of Even and Odd Terms	Sum of even terms = $\frac{(1 + 1)^{n} + (1 - 1)^{n}}{2} = 2^{n - 1}$ ; sum of odd terms = $\frac{(1 + 1)^{n} - (1 - 1)^{n}}{2} = 2^{n - 1}$ .	When asked for the sum of coefficients of terms with even powers or odd powers.
Using Binomial Theorem for Divisibility	Example: $6^{n} - 5 n - 1$ is divisible by 25. Write $6 = 1 + 5$ , so $6^{n} = 1 + 5 n + 25 (\dots)$ .	When proving that a large number is divisible by another, e.g., $7^{2 n} - 1$ divisible by 48.
Multinomial Theorem Extension	$(x_{1} + x_{2} + \dots + x_{m})^{n} = \sum \frac{n!}{r_{1}! r_{2}! \dots r_{m}!} x_{1}^{r_{1}} x_{2}^{r_{2}} \dots x_{m}^{r_{m}}$ where $\sum r_{i} = n$ .	When expanding expressions like $(a + b + c)^{n}$ or finding coefficients in such expansions.
Particular Term from the End	Term from end: $T_{n - r + 2} = (\binom{n}{n - r + 1}) a^{r - 1} b^{n - r + 1}$	When asked for a term near the end of the expansion, e.g., 3rd term from the end.
Binomial Theorem for Negative/Fractional Index	$(1 + x)^{- n} = \sum_{r = 0}^{\infty} (- 1)^{r} (\binom{n + r - 1}{r}) x^{r}$	When expanding $1 / (1 + x)^{n}$ or similar expressions.
Finding Numerically Greatest Term in $(a + b)^{n}$	Let $x = b / a$ . Then $\frac{T_{r + 1}}{T_{r}} = \frac{n - r + 1}{r} ∣ x ∣$ . Solve $\frac{n - r + 1}{r} ∣ x ∣ \geq 1$ .	When asked for the numerically greatest term in $(a + b)^{n}$ .
Pascal's Triangle	$(\binom{n}{r}) = (\binom{n - 1}{r - 1}) + (\binom{n - 1}{r})$	To quickly generate binomial coefficients for small $n$ without factorials.
Binomial Theorem in Series Summation	$\sum_{r = 0}^{n} r (\binom{n}{r}) = n 2^{n - 1}$ (from $x = 1$ after differentiating $n (1 + x)^{n - 1}$ ).	When evaluating sums involving $r$ times binomial coefficients.
Binomial Theorem for Product of Two Binomials	Coefficient of $x^{m}$ in $(1 + x)^{n} (1 + x)^{p} = \sum_{r = 0}^{m} (\binom{n}{r}) (\binom{p}{m - r})$	When finding coefficients in the product of two binomial expansions.
Binomial Theorem for Negative Exponents with Two Terms	$(a + b x)^{- n} = a^{- n} \sum_{r = 0}^{\infty} (- 1)^{r} (\binom{n + r - 1}{r}) (b / a)^{r} x^{r}$	When expanding expressions like $(2 - 3 x)^{- 4}$ .
Finding Remainder Using Binomial Theorem	Example: $2^{100} = (1 + 1)^{100} = \sum (\binom{100}{r})$ , remainder mod 7? Write $2^{3} = 8 = 7 + 1$ , so $2^{100} = 2^{99} \cdot 2 = (2^{3})^{33} \cdot 2 = (7 + 1)^{33} \cdot 2$ , remainder $1^{33} \cdot 2 = 2$ .	When finding remainder of large powers divided by small numbers.
Binomial Theorem for Approximating Roots	$\sqrt[n]{1 + x} = (1 + x)^{1 / n} = 1 + \frac{1}{n} x + \frac{1 / n (1 / n - 1)}{2!} x^{2} + \dots$	When approximating square roots, cube roots, etc., like $\sqrt{1.04}$ .

Open the Binomial Theorem formula page

What JEE asks

Binomial Theorem Statement. JEE often asks for the coefficient of a specific term, e.g., $x^{5}$ in $(2 x - 3)^{7}$ , requiring identification of $r$ .
General Term in Binomial Expansion. JEE frequently tests finding the term independent of $x$ or the coefficient of a given power by setting the exponent of $x$ to zero or the desired value.
Finding Term Independent of x. JEE often combines this with rationalizing or simplifying expressions like $(\sqrt{x} + 1 / \sqrt[3]{x})^{n}$ .
Middle Term(s) in Binomial Expansion. JEE may ask for the greatest coefficient among middle terms, requiring comparison of $(\binom{n}{r})$ values.
Binomial Coefficient Properties. JEE often uses these to evaluate series like $\sum_{r = 0}^{n} r (\binom{n}{r})$ by differentiating $(1 + x)^{n}$ .
Greatest Term in Binomial Expansion. JEE may ask for the greatest coefficient (when $x = 1$ ) or greatest term for a specific $x$ .
Binomial Theorem for Any Index. JEE uses this for approximations and expansions like $(1 + x)^{- 1}$ or $(1 - x)^{- 2}$ .
Approximations Using Binomial Theorem. JEE often asks for the approximate value of expressions like $(1.02)^{7}$ or $(0.98)^{- 4}$ .
Finding Coefficient of a Specific Power. JEE often combines this with multiple binomials or products, requiring careful algebra.
Binomial Coefficient Notation. JEE tests understanding of properties like $(\binom{n}{r}) = (\binom{n}{n - r})$ and Pascal's rule.
Sum of Coefficients in Binomial Expansion. JEE often asks for sum of coefficients of even/odd terms using $x = 1$ and $x = - 1$ .
Sum of Coefficients of Even and Odd Terms. JEE may ask for the sum of coefficients of terms where the exponent is a multiple of 3, using complex cube roots of unity.
Using Binomial Theorem for Divisibility. JEE often asks for the remainder when a large power is divided by a small number, e.g., $2^{100}$ mod 7.
Multinomial Theorem Extension. JEE may ask for the coefficient of $x^{2} y^{3} z$ in $(x + y + z)^{6}$ .
Particular Term from the End. JEE may combine this with finding the coefficient of a specific power.
Binomial Theorem for Negative/Fractional Index. JEE may ask for the coefficient of $x^{r}$ in $(1 - x)^{- 3}$ using the formula.
Finding Numerically Greatest Term in $(a + b)^{n}$ . JEE often tests this with specific numbers like $(2 + 3 x)^{7}$ for a given $x$ .
Pascal's Triangle. JEE may ask for the sum of elements in a row or diagonal properties.
Binomial Theorem in Series Summation. JEE often asks for $\sum r (r - 1) (\binom{n}{r})$ using second derivative.
Binomial Theorem for Product of Two Binomials. JEE may ask for coefficient of $x^{5}$ in $(1 + x)^{7} (1 + x)^{8}$ or similar.
Binomial Theorem for Negative Exponents with Two Terms. JEE may ask for the first few terms or the general term.
Finding Remainder Using Binomial Theorem. JEE often asks for remainder when $2^{100}$ is divided by 7, or similar.
Binomial Theorem for Approximating Roots. JEE may ask for the approximate value of $\sqrt{101}$ by writing $\sqrt{100 (1 + 0.01)}$ .

Common traps

Forgetting that the sum runs from $r = 0$ to $n$ , not $1$ to $n$ . Binomial Theorem Statement
Confusing $T_{r + 1}$ with $T_{r}$ ; the index starts at $r = 0$ for the first term. General Term in Binomial Expansion
Forgetting to consider the constant term from both factors; sometimes the exponent equation yields a non-integer $r$ (no independent term). Finding Term Independent of x
Using $n / 2$ as the term number instead of $n / 2 + 1$ . Middle Term(s) in Binomial Expansion
Confusing $\sum (\binom{n}{r}) = 2^{n}$ with $\sum {(\binom{n}{r})}^{2} = (\binom{2 n}{n})$ . Binomial Coefficient Properties
Forgetting to consider absolute value of $x$ when $x$ is negative. Greatest Term in Binomial Expansion
Applying this for $∣ x ∣ \geq 1$ without convergence considerations. Binomial Theorem for Any Index
Using the approximation when $x$ is not small (e.g., $x > 0.1$ leads to large error). Approximations Using Binomial Theorem
Forgetting to consider the coefficient from $a$ and $b$ as well. Finding Coefficient of a Specific Power
Thinking $(\binom{n}{r}) = \frac{n!}{r!}$ (missing the $(n - r)!$ factor). Binomial Coefficient Notation
Forgetting that this works only for $(1 + x)^{n}$ ; for $(a + b x)^{n}$ , substitute $x = 1$ after factoring. Sum of Coefficients in Binomial Expansion
Assuming the sum of even and odd terms are different; they are equal for $(1 + x)^{n}$ . Sum of Coefficients of Even and Odd Terms
Not expanding fully to see the remainder pattern. Using Binomial Theorem for Divisibility
Forgetting that the sum of exponents must equal $n$ . Multinomial Theorem Extension

Topics in Binomial Theorem

Binomial Expansion
General and Middle Term
Binomial Coefficients
Properties of Coefficients
Applications of Binomial Theorem

Previous-year questions on Binomial Theorem

JEEGenie has 350 previous-year questions on Binomial Theorem. A few worked examples follow.

Year	Questions
2026	17
2025	25
2024	23
2023	47
2022	29
2021	40
2020	20
2019	40
2018	9
2017	5
2016	7
2015	5
2014	7
2013	6
2012	4
2011	1
2010	4
2009	1
2008	1
2007	2
2006	2
2005	5
2004	6
2003	6
2002	6
2001	1
2000	3
1999	1
1998	2
1997	2
1994	3
1993	2
1992	4
1989	1
1986	2
1985	2
1983	3
1982	2
1980	1
1979	1
1978	2

Question 1 medium

If for

3 \leq r \leq 30

(^{30} C_{30 - r}) + 3 (^{30} C_{31 - r}) + 3 (^{30} C_{32 - r}) + (^{30} C_{33 - r}) =^{m} C_{r}

, then m equals :

Answer: C

Explanation.

^{30} C_{30 - r} + 3^{30} C_{31 - r} + 3^{30} C_{32 - r} +^{30} C_{33 - r}

Let

k = 30 - r

. Then the expression becomes

^{30} C_{k} + 3^{30} C_{k + 1} + 3^{30} C_{k + 2} +^{30} C_{k + 3} .

Group using Pascal’s identity:

(^{30} C_{k} +^{30} C_{k + 1}) + 2 (^{30} C_{k + 1} +^{30} C_{k + 2}) + (^{30} C_{k + 2} +^{30} C_{k + 3})

=^{31} C_{k + 1} + 2^{31} C_{k + 2} +^{31} C_{k + 3} .

Group again:

(^{31} C_{k + 1} +^{31} C_{k + 2}) + (^{31} C_{k + 2} +^{31} C_{k + 3})

=^{32} C_{k + 2} +^{32} C_{k + 3} .

Apply Pascal’s identity once more:

^{32} C_{k + 2} +^{32} C_{k + 3} =^{33} C_{k + 3} .

Now

k + 3 = (30 - r) + 3 = 33 - r

. Hence

^{33} C_{33 - r} =^{33} C_{r} .

Thus the given expression equals

^{33} C_{r}

, so

m = 33

33

Option C.

From NTA/IIT papers: JEE Main 2026 · 2 Apr Shift 2

Question 2 hard

The term independent of

x

in the expansion of

{(\frac{(x + 1)}{(x^{2 / 3} + 1 - x^{1 / 3})} - \frac{(x - 1)}{(x - x^{1 / 2})})}^{10}, x > 1

, is :

Answer: D

Explanation. First simplify each fraction. For the first term:

x^{2 / 3} + 1 - x^{1 / 3} = (x^{1 / 3})^{2} - x^{1 / 3} + 1

. Note

x + 1 = (x^{1 / 3} + 1) (x^{2 / 3} - x^{1 / 3} + 1)

. Thus

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

. For the second term:

x - x^{1 / 2} = x^{1 / 2} (x^{1 / 2} - 1)

. Also

x - 1 = (x^{1 / 2} - 1) (x^{1 / 2} + 1)

. Thus

\frac{x - 1}{x - x^{1 / 2}} = \frac{x^{1 / 2} + 1}{x^{1 / 2}} = 1 + x^{- 1 / 2}

. So the expression becomes:

(x^{1 / 3} + 1 - (1 + x^{- 1 / 2}))^{10} = (x^{1 / 3} - x^{- 1 / 2})^{10}

. General term:

T_{r + 1} = (\binom{10}{r}) (x^{1 / 3})^{10 - r} (- x^{- 1 / 2})^{r} = (\binom{10}{r}) (- 1)^{r} x^{(10 - r) / 3 - r / 2}

. Exponent of

x

\frac{10 - r}{3} - \frac{r}{2} = \frac{20 - 2 r - 3 r}{6} = \frac{20 - 5 r}{6}

. Set

20 - 5 r = 0 \Rightarrow r = 4

. Term independent of

x

T_{5} = (\binom{10}{4}) (- 1)^{4} = (\binom{10}{4}) = 210

. Thus answer is option D.

From NTA/IIT papers: JEE Main 2025 · 2 Apr Shift 1

Question 3 easy

If A denotes the sum of all the coefficients in the expansion of

{(1 - 3 x + 10 x^{2})}^{n}

and B denotes the sum of all the coefficients in the expansion of

{(1 + x^{2})}^{n}

, then :

$B = A^{3}$
$3 A = B$
$A = 3 B$
$A = B^{3}$

Answer: D

Explanation. The sum of coefficients in a polynomial expansion is found by substituting

x = 1

. For

A

A = (1 - 3 (1) + 10 (1)^{2})^{n} = (1 - 3 + 10)^{n} = 8^{n}

. For

B

B = (1 + 1^{2})^{n} = (2)^{n} = 2^{n}

. Now

8^{n} = (2^{3})^{n} = (2^{n})^{3} = B^{3}

. Thus

A = B^{3}

. The correct answer is option D.

From NTA/IIT papers: JEE Main 2024 · 27 Jan Shift 1

Question 4 medium

^{2 n} C_{3} :^{n} C_{3} = 10 : 1

, then the ratio

(n^{2} + 3 n) : (n^{2} - 3 n + 4)

is :

$27 : 11$
$2 : 1$
$35 : 16$
$65 : 37$

Answer: B

Explanation. Given

^{2 n} C_{3} :^{n} C_{3} = 10 : 1

, we have:

\frac{\frac{2 n (2 n - 1) (2 n - 2)}{6}}{\frac{n (n - 1) (n - 2)}{6}} = 10

Simplify:

\frac{2 n (2 n - 1) (2 n - 2)}{n (n - 1) (n - 2)} = 10

Factor:

2 n - 2 = 2 (n - 1)

, so:

\frac{2 n (2 n - 1) \cdot 2 (n - 1)}{n (n - 1) (n - 2)} = \frac{4 (2 n - 1)}{n - 2} = 10

Thus:

4 (2 n - 1) = 10 (n - 2) ⟹ 8 n - 4 = 10 n - 20 ⟹ 2 n = 16 ⟹ n = 8

Now compute ratio:

n^{2} + 3 n = 64 + 24 = 88, n^{2} - 3 n + 4 = 64 - 24 + 4 = 44

Ratio

88 : 44 = 2 : 1

. Thus answer is option B.

From NTA/IIT papers: JEE Main 2023 · 6 Apr Shift 1

Question 5 hard

Let for the

9^{th}

term in the binomial expansion of

(3 + 6 x)^{n}

, in the increasing powers of

6 x

, to be the greatest for

x = \frac{3}{2}

, the least value of

n

n_{0}

. If

k

is the ratio of the coefficient of

x^{6}

to the coefficient of

x^{3}

, then

k + n_{0}

is equal to :

Answer: 24

Explanation. **Principle:** In

(3 + 6 x)^{n}

the ratio of consecutive terms is

\frac{T_{r + 1}}{T_{r}} = \frac{n - r + 1}{r} \cdot \frac{6 x}{3} .

x = \frac{3}{2}

\frac{6 x}{3} = 3

, so

\frac{T_{r + 1}}{T_{r}} = \frac{3 (n - r + 1)}{r}

. **9th term greatest:** need

T_{9} \geq T_{8}

and

T_{9} \geq T_{10}

T_{9} \geq T_{8}

(

r = 8

\frac{3 (n - 7)}{8} \geq 1 \Rightarrow n \geq 7 + \frac{8}{3} = \frac{29}{3} \approx 9.67 T_{10} \leq T_{9}

(

r = 9

\frac{3 (n - 8)}{9} \leq 1 \Rightarrow n \leq 11

\frac{29}{3} \leq n \leq 11

, and the least integer is

n_{0} = 10

. **Ratio of coefficients:**

(3 + 6 x)^{10} = 3^{10} (1 + 2 x)^{10}

. Coefficient of

x^{6}

3^{10} (\binom{10}{6}) 2^{6}

; coefficient of

x^{3}

3^{10} (\binom{10}{3}) 2^{3}

k = \frac{(\binom{10}{6}) 2^{6}}{(\binom{10}{3}) 2^{3}} = \frac{210 \times 64}{120 \times 8} = 14

k + n_{0} = 14 + 10 = 24

Answer: 24

From NTA/IIT papers: JEE Main 2022 · 27 Jul Shift 2

Frequently asked questions

How much of JEE Main comes from Binomial Theorem?: JEEGenie estimates about 3% of JEE Main marks come from Binomial Theorem, based on recent papers. The share varies from year to year.
Which formulas matter most in Binomial Theorem?: The Binomial Theorem sheet on this page has 23 formula entries, starting with Binomial Theorem Statement, General Term in Binomial Expansion, Finding Term Independent of x, Middle Term(s) in Binomial Expansion, Binomial Coefficient Properties. Each one comes with when to use it and the trap to avoid.
What topics does Binomial Theorem cover for JEE?: Binomial Theorem covers Binomial Expansion, General and Middle Term, Binomial Coefficients, Properties of Coefficients, Applications of Binomial Theorem.
How should I revise Binomial Theorem before the exam?: Learn each formula together with its condition of use, read the trap listed beside it, then practise questions and review every mistake. The practice link on this page opens Binomial Theorem questions on JEEGenie.
How many previous-year questions are there on Binomial Theorem?: JEEGenie has 350 previous-year questions on Binomial Theorem, and this page shows a few with worked explanations.
Is practice on Binomial Theorem free on JEEGenie?: Sign-up is free and includes a 7-day Pro trial. After the trial, free accounts keep a limited daily quota.