Divisibility Rules of Binomial Expansion

An algebraic expression consisting of only two terms is called a Binomial Expression. Expressions with higher powers are difficult to solve. In these cases, Binomial theorem plays a major role. Binomial theorem is used for the expansion of a binomial expression with a higher degree. Binomial theorem is proved using the concept of mathematical induction. Apart from mathematics, Binomial theorem is also used in various fields for statistical and financial data analysis.

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This article is about the divisibility rules of binomial expansion which falls under the topic Binomial theorem and its applications. This is one of the important topics for competitive exams.

Binomial Theorem

If $n$ is any positive integer, then

$ (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 - 1} a b^{n - 1} + \binom{n}{n} b^n $

Special cases of Binomial Theorem:

In the expansion of $ (a + b)^n $,

(i) Taking $ a = x $ and $ b = -y $, we obtain:

$(x - y)^n = [x + (-y)]^n $

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

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

Thus,

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

Using this, we have:

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

(ii) Taking $ a = 1 $, $ b = x $, we obtain:

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

$ = \binom{n}{0} + \binom{n}{1} x + \binom{n}{2} x^2 + \binom{n}{3} x^3 + \dots + \binom{n}{n} x^n $

Thus,

$ (1 + x)^n = \binom{n}{0} + \binom{n}{1} x + \binom{n}{2} x^2 + \binom{n}{3} x^3 + \dots + \binom{n}{n} x^n $

In particular, for $ x = 1 $, we have:

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

(iii) Taking $ a = 1 $, $ b = -x $, we obtain:

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

In particular, for $ x = 1 $, we get:

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

Divisibility Rules:

To find whether the binomial expansion is divisible by an integer,

1. Expression, $(1+x)^{\mathrm{n}}-1$ is divisible by x because

$ (1+\mathrm{x})^{\mathrm{n}}-1 =\left({ }^{\mathrm{n}} \mathrm{C}_0+{ }^{\mathrm{n}} \mathrm{C}_1 \mathrm{x}+{ }^{\mathrm{n}} C_2 x^2+\ldots . .+{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}} \mathrm{x}^{\mathrm{n}}\right)-1 $

$ =\mathrm{x}\left[{ }^{\mathrm{n}} \mathrm{C}_1+{ }^{\mathrm{n}} C_2 \mathrm{x}+\ldots \ldots .+{ }^{\mathrm{n}} C_{\mathrm{n}} \mathrm{x}^{\mathrm{n}-1}\right] $

2. Expression, $(1+\mathrm{x})^{\mathrm{n}}-\mathrm{nx}-1$ is divisible by $\mathrm{x}^2$ because

$ (1+\mathrm{x})^{\mathrm{n}}-\mathrm{nx}-1=\left({ }^{\mathrm{n}} \mathrm{C}_0+{ }^{\mathrm{n}} \mathrm{C}_1 \mathrm{x}+{ }^{\mathrm{n}} \mathrm{C}_2 \mathrm{x}^2+\ldots \ldots+{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}} \mathrm{x}^{\mathrm{n}}\right)-\mathrm{nx}-1 $

$ =x^2\left[{ }^n C_2+{ }^{\mathrm{n}} \mathrm{C}_3 \mathrm{x}+\ldots \ldots+{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}} \mathrm{x}^{\mathrm{n}-2}\right] $

For example:

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

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

$ =\left[1+(n+1) 8+{ }^{n+1} C_2 8^2+\ldots\right]-8 n-9$

$ ={ }^{n+1} C_2 8^2+{ }^{n+1} C_3 8^3+{ }^{n+1} C_4 8^4+\ldots $

$ =8\left[{ }^{n+1} C_2 8+{ }^{n+1} C_3 8^2+{ }^{n+1} C_4 8^3+\ldots\right] $

which is clearly divisible by 8 .

1. The expression $a^{\mathrm{n}}-\mathrm{b}^{\mathrm{n}}$ is divisible by $\mathrm{a}+\mathrm{b}$, if n is even.

2. The expression $a^n-b^n$ is divisible by $a-b$, if $n$ is even or odd.

3. The expression $a^{\mathrm{n}}+b^{\mathrm{n}^i}$ divisible by, $\mathrm{a}+$ bif n is odd.

In all the above cases $n$ is a natural number.

For Example,

The expression $15^4-7^4$ is divisible by $(15+7)=22$ and $(15-7)=8$

Recommended Video Based on Divisibility Rules of Binomial Expansion:

Solved Examples based on divisibility rules of binomial expansion:

Example 1: Which of the following is true for the number $2^{4 n}-2^n(7 n+1)$, 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^{4 n}-2^n(7 n+1)=(16)^n-2^n(7 n+1) $

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

$ \quad=\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 7 n-2^n $

$ \quad=14^2\left({ }^n C_2 2^{n-2}+{ }^n C_3 2^{n-3} 14+\ldots+14^{n-2}\right)+\left(2^n+{ }^n C_1 \cdot 2^{n-1} \cdot 14-2^n \cdot 7 n-2^n\right) $

$ \quad=14^2\left({ }^n C_2 2^{n-2}+{ }^n C_3 2^{n-3} \cdot 14+\ldots+14^{n-2}\right)+\left(2^n+n 2^{n-1} \cdot 2^1 \cdot 7-2^n \cdot 7 n-2^n\right) $

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

This is divisible by $14^2$

Hence, the answer is the 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} \mathrm{C}_0+{ }^{100} \mathrm{C}_1(100)+\ldots\right)-1 $

$ =10000\left(1+{ }^{100} \mathrm{C}_2+\ldots\right) $

obviously 10000 is the correct choice.

Hence, the answer is the option (3).

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

1) $2 x$

2) $x^2$

3) $2 x^3$

4) All of these

Solution:

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

$ \Rightarrow(1+x)^n-n x-1 $

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

From above it is clear that $(1+x)^n-n x-1$ is divisible by $x^2$.

Trick: $(1+x)^n-n x-1$. Put $\mathrm{n}=2$ and $\mathrm{x}=3$; Then $4^2-2.3-1=9$ is not divisible by 6,54 but divisible by 9 . Which is given by option (b) i.e., $x^2=9$

Hence, the answer is the option (2).

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

1) 3

2) 19

3) 64

4) 29

Solution:

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

$ 1+{ }^n C_1(48)+{ }^n C_2(48)^2+\ldots+{ }^n C_n(48)^n+16 n-1$

$= (48 n+16 n)+{ }^n C_2(48)^2+{ }^n C_3(48)^3+\ldots+{ }^n C_n(48)^n $

$= 64 n+8^2\left[{ }^n C_2 \cdot 6^2+{ }^n C_3 \cdot 6^3 \cdot 8+{ }^n C_4 \cdot 6^4 \cdot 8^2+\ldots+{ }^n C_n \cdot 6^n \cdot 8^{n-2}\right] $

Hence, $49^n+16 n-1$ is divisible by 64 .

Hence, the answer is option (3).

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

1) 0

2) 1

3) 2

4) None of these

Solution:

$ 7^{2 n}+2^{3 n-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}$

$ =\text { a multiple of } 25, \text { since }(-1)^n+(-1)^{n-1}=0 $

Hence, the answer is option (1).