Divisibility Rules of Binomial Expansion

Binomial Theorem

If $n$ is any positive integer, then the Binomial Theorem states:

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

This expansion is used throughout Class 11 and competitive exams to simplify algebraic expressions and understand patterns in binomial coefficients.

Special Cases of the Binomial Theorem

This section covers the important variations of the Binomial Theorem that you frequently use in Class 11. It explains how the theorem changes when you replace $b$ with $-y$, or take specific values like $a = 1$ or $x = 1$.

Expansion of $(x - y)^n$ Using the Binomial Theorem

Let $a = x$ and $b = -y$. Then:

$(x - y)^n = \binom{n}{0}x^n - \binom{n}{1}x^{n-1}y + \binom{n}{2}x^{n-2}y^2 - \dots + (-1)^n \binom{n}{n}y^n$

This form is extremely useful for dealing with alternating signs and evaluating expressions involving differences of powers.

Example

Expand $(x - 2y)^5$:

$(x - 2y)^5 = \binom{5}{0}x^5 - \binom{5}{1}x^4(2y) + \binom{5}{2}x^3(2y)^2 - \binom{5}{3}x^2(2y)^3 + \binom{5}{4}x(2y)^4 - \binom{5}{5}(2y)^5$

$= x^5 - 10x^4y + 40x^3y^2 - 80x^2y^3 + 80xy^4 - 32y^5$

Expansion of $(1 + x)^n$

If $a = 1$ and $b = x$:

$(1 + x)^n = \binom{n}{0} + \binom{n}{1}x + \binom{n}{2}x^2 + \dots + \binom{n}{n}x^n$

Important Result

For $x = 1$:

$2^n = \binom{n}{0} + \binom{n}{1} + \binom{n}{2} + \dots + \binom{n}{n}$

This identity is used frequently in binomial coefficient sums and divisibility problems.

Expansion of $(1 - x)^n$

If $a = 1$ and $b = -x$:

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

Important Result

For $x = 1$:

$0 = \binom{n}{0} - \binom{n}{1} + \binom{n}{2} - \dots + (-1)^n \binom{n}{n}$

This alternating-sum identity is a fundamental part of binomial divisibility problems.

Divisibility Rules in Binomial Expansion

Divisibility rules help determine whether an expression involving binomial powers is divisible by certain integers. These shortcuts save time in exams and strengthen number-system reasoning.

Rule 1 — $(1 + x)^n - 1$ is Divisible by $x$

$(1 + x)^n - 1 = \left(\binom{n}{0} + \binom{n}{1}x + \binom{n}{2}x^2 + \dots + \binom{n}{n}x^n\right) - 1$

$= x\left(\binom{n}{1} + \binom{n}{2}x + \dots + \binom{n}{n}x^{n-1}\right)$

This proves that $(1 + x)^n - 1$ always has a factor of $x$.

Rule 2 — $(1 + x)^n - nx - 1$ is Divisible by $x^2$

$(1 + x)^n - nx - 1 = \left(\binom{n}{0} + \binom{n}{1}x + \binom{n}{2}x^2 + \dots\right) - nx - 1$

$= x^2\left(\binom{n}{2} + \binom{n}{3}x + \dots + \binom{n}{n}x^{n-2}\right)$

Hence, $(1 + x)^n - nx - 1$ always contains $x^2$ as a factor.

Example — Proving Divisibility by 8

Prove that $3^{2n+2} - 8n - 9$ is divisible by $8$ for $n \in \mathbb{N}$.

Rewrite:

$3^{2n+2} - 8n - 9 = (1 + 8)^{,n+1} - 8n - 9$

Expand using Binomial Theorem:

$(1 + 8)^{n+1} = 1 + (n+1)8 + \binom{n+1}{2}8^2 + \binom{n+1}{3}8^3 + \dots$

Subtract the linear part:

$= \binom{n+1}{2}8^2 + \binom{n+1}{3}8^3 + \dots$

Factor out $8$:

$= 8\left(\binom{n+1}{2}8 + \binom{n+1}{3}8^2 + \dots\right)$

Clearly divisible by $8$.

Divisibility Rules for Expressions Involving Powers

This section explains the key divisibility properties of expressions like $a^n - b^n$ and $a^n + b^n$. These rules help you quickly determine when such expressions are divisible by $a+b$, $a-b$, or other factors, making them especially useful for solving Class 11 algebra and number system problems without long calculations.

Rule 1 — $a^n - b^n$ is Divisible by $a + b$ (when $n$ is even)

If $n$ is even, then
$ a^n - b^n = (a - b)(a^{n-1} + a^{n-2}b + \ldots + b^{,n-1}) $
and this expression can also be rewritten as:
$ a^n - b^n = (a + b)(a^{n-1} - a^{n-2}b + \ldots - b^{,n-1}) $
because the signs alternate when $n$ is even.
So $(a + b)$ becomes a clear factor.

Example: $4^2 - 2^2 = 16 - 4 = 12$, and $12$ is divisible by $(4+2)=6$.

Rule 2 — $a^n - b^n$ is Always Divisible by $a - b$ (for any $n$)

For every natural number $n$:
$ a^n - b^n = (a - b)(a^{n-1} + a^{n-2}b + \ldots + b^{,n-1}) $
This factorization works for both even and odd values of $n$.

Example: $7^3 - 4^3 = 343 - 64 = 279$, and $279$ is divisible by $(7-4)=3$.

Rule 3 — $a^n + b^n$ is Divisible by $a + b$ (when $n$ is odd)

For odd $n$,
$ a^n + b^n = (a + b)(a^{n-1} - a^{n-2}b + \ldots - b^{,n-1}) $
The alternating signs appear only when $n$ is odd, giving $(a + b)$ as a factor.

Example: $2^3 + 1^3 = 8 + 1 = 9$, and $9$ is divisible by $(2+1)=3$.

Example

The expression $15^4 - 7^4$ is divisible by:

• $(15 + 7) = 22$, because $n = 4$ is even

• $(15 - 7) = 8$, always a factor of $a^n - b^n$

Why These Divisibility Rules Matter

These rules make it easier to:

• solve binomial remainder questions

• compute last digits

• simplify expressions involving large powers

• handle Class 11 NCERT exercises effortlessly

• prepare for JEE, Olympiads, and competitive exams

They provide shortcuts that eliminate the need for full expansion, making calculations faster and more accurate.

Solved Examples based on divisibility rules of binomial expansion:

Example 1:
Which of the following is true for the number $2^{4n} - 2^n(7n + 1)$, where $n$ is any positive integer?

1. The highest divisor of the number among the given options is $7$.

2. The highest divisor of the number among the given options is $72$.

3. The highest divisor of the number among the given options is $142$.

4. The highest divisor of the number among the given options is $14$.

Solution:

$2^{4n} - 2^n(7n + 1) = (16)^n - 2^n(7n + 1)$

$= (2 + 14)^n - 2^n \cdot 7n - 2^n$

$= \left(2^n + {}^n C_1 2^{n-1} \cdot 14 + {}^n C_2 2^{n-2} \cdot 14^2 + \ldots + 14^n \right) - 2^n \cdot 7n - 2^n$

$= 14^2\left({}^n C_2 2^{n-2} + {}^n C_3 2^{n-3} \cdot 14 + \ldots + 14^{n-2}\right)$

Thus, the expression is divisible by $14^2 = 196$.
Among the options, the corresponding choice is 142.

Hence, the answer is option 3.

Example 2:
The number $101^{100} - 1$ is divisible by:

1. $101$

2. $1001$

3. $10000$

4. $100000$

Solution:

$(101)^{100} - 1 = (1 + 100)^{100} - 1$

$= \left({}^{100}C_0 + {}^{100}C_1(100) + \ldots \right) - 1$

$= 10000\left(1 + {}^{100}C_2 + \ldots \right)$

Clearly, the expression is divisible by $10000$.

Hence, the answer is option 3.

Example 3:
$(1 + x)^n - nx - 1$ is divisible by (where $n \in \mathbb{N}$):

1. $2x$

2. $x^2$

3. $2x^3$

4. All of these

Solution:

$(1 + x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \ldots$

So,

$(1 + x)^n - nx - 1 = x^2\left[\frac{n(n-1)}{2!} + \frac{n(n-1)(n-2)}{3!}x + \ldots\right]$

Thus, the expression is divisible by $x^2$.

Hence, the answer is option 2.

Example 4:
$49^n + 16n - 1$ is divisible by:

1. $3$

2. $19$

3. $64$

4. $29$

Solution:

$49^n + 16n - 1 = (1 + 48)^n + 16n - 1$

$= 1 + {}^nC_1(48) + {}^nC_2(48)^2 + \ldots + 16n - 1$

$= (48n + 16n) + {}^nC_2(48)^2 + {}^nC_3(48)^3 + \ldots$

$= 64n + 8^2\left[{}^nC_2 \cdot 6^2 + {}^nC_3 \cdot 6^3 \cdot 8 + \ldots\right]$

Clearly, the expression contains a factor of $64$.

Hence, the answer is option 3.

Example 5:
The greatest integer $m$ such that $5^m$ divides $7^{2n} + 2^{3n-3} \cdot 3^{n-1}$ for $n \in \mathbb{N}$ is:

1. $0$

2. $1$

3. $2$

4. None of these

Solution:

$7^{2n} + 2^{3n-3} \cdot 3^{n-1} = 49^n + 24^{n-1}$

$= (50 - 1)^n + (25 - 1)^{n-1}$

$= \text{multiple of } 50 + (-1)^n + \text{multiple of } 25 + (-1)^{n-1}$

Since $(-1)^n + (-1)^{n-1} = 0$,
the entire expression is a multiple of $25 = 5^2$, but does not guarantee $5^3$.

So the highest guaranteed power is $5^0$.

Hence, the answer is option 1.

List of Topics Related to Divisibility Rules of Binomial Expansion

This section gives you a quick overview of the key topics connected to divisibility in binomial expansion. It highlights important ideas such as the Binomial Theorem for any index, methods to identify the greatest binomial coefficient, essential results used for divisibility checks, and techniques for finding last digits and remainders using binomial expansion. It serves as a compact guide to what you need to study for Class 11 and exam-level questions.

Binomial Theorem for any Index

Greatest Binomial Coefficient

Important Results of Binomial Theorem for any Index

Last Digits and Remainder using the Binomial Expansion

NCERT Resources

This section provides all the NCERT-based study resources you need to understand the divisibility rules in binomial expansion. You’ll find clear Class 11 chapter notes, step-by-step NCERT solutions, and exemplar problems that help strengthen concepts and improve accuracy for school exams and competitive tests.

NCERT Maths Class 11 Notes for Chapter 8 - Binomial Theorem and its applications

NCERT Maths Class 11 Solutions for Chapter 8 - Binomial Theorem and its applications

NCERT Maths Class 11 Exemplar Solutions for Chapter 8 - Binomial Theorem and its applications

Practice Questions based on the Divisibility Rules of Binomial Expansion

This section offers practice questions designed to help you apply divisibility rules within binomial expansion. These problems reinforce key identities, test your understanding of coefficients and remainders, and prepare you for the type of questions commonly asked in Class 11 exams and competitive tests.

Problems On Divisibility- Practice Question MCQ

We have shared below the links to practice questions for the topics related to the binomial theorem: