Divisibility Rules of Binomial Expansion

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).