Careers360 Logo
Divisibility Rules of Binomial Expansion

Divisibility Rules of Binomial Expansion

Edited By Komal Miglani | Updated on Feb 11, 2025 08:04 PM IST

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.

Divisibility Rules of Binomial Expansion
Divisibility Rules of Binomial Expansion

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=(n0)an+(n1)an1b+(n2)an2b2++(nn1)abn1+(nn)bn

Special cases of Binomial Theorem:

In the expansion of (a+b)n,

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

(xy)n=[x+(y)]n

=(n0)xn+(n1)xn1(y)+(n2)xn2(y)2+(n3)xn3(y)3++(nn)(y)n

=(n0)xn(n1)xn1y+(n2)xn2y2(n3)xn3y3++(1)n(nn)yn

Thus,

(xy)n=(n0)xn(n1)xn1y+(n2)xn2y2++(1)n(nn)yn

Using this, we have:

(x2y)5=(50)x5(51)x4(2y)+(52)x3(2y)2(53)x2(2y)3+(54)x(2y)4(55)(2y)5

=x510x4y+40x3y280x2y3+80xy432y5

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

(1+x)n=(n0)(1)n+(n1)(1)n1x+(n2)(1)n2x2++(nn)xn

=(n0)+(n1)x+(n2)x2+(n3)x3++(nn)xn

Thus,

(1+x)n=(n0)+(n1)x+(n2)x2+(n3)x3++(nn)xn

In particular, for x=1, we have:

2n=(n0)+(n1)+(n2)++(nn)

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

(1x)n=(n0)(n1)x+(n2)x2+(1)n(nn)xn

In particular, for x=1, we get:

0=(n0)(n1)+(n2)+(1)n(nn)

Divisibility Rules:

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

1. Expression, (1+x)n1 is divisible by x because

(1+x)n1=(nC0+nC1x+nC2x2+..+nCnxn)1

=x[nC1+nC2x+.+nCnxn1]

2. Expression, (1+x)nnx1 is divisible by x2 because

(1+x)nnx1=(nC0+nC1x+nC2x2++nCnxn)nx1

=x2[nC2+nC3x++nCnxn2]

For example:

Prove that 32n+28n9 is divisible by 8 if nN

32n+28n9=(1+8)n+18n9

=[1+(n+1)8+n+1C282+]8n9

=n+1C282+n+1C383+n+1C484+

=8[n+1C28+n+1C382+n+1C483+]

which is clearly divisible by 8 .

1. The expression anbn is divisible by a+b, if n is even.

2. The expression anbn is divisible by ab, if n is even or odd.

3. The expression an+bni divisible by, a+ bif n is odd.

In all the above cases n is a natural number.

For Example,

The expression 15474 is divisible by (15+7)=22 and (157)=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 24n2n(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:

24n2n(7n+1)=(16)n2n(7n+1)

=(2+14)n2n7n2n

=(2n+nC12n114+nC22n2142++14n)2n7n2n

=142(nC22n2+nC32n314++14n2)+(2n+nC12n1142n7n2n)

=142(nC22n2+nC32n314++14n2)+(2n+n2n12172n7n2n)

=142(nC22n2+nC32n314++14n2)

This is divisible by 142

Hence, the answer is the option 3.

Example 2: The number 1011001 is divisible by

1) 101

2) 1001

3) 10000

4) 100000

Solution:

(101)1001

=(1+100)1001

=(100C0+100C1(100)+)1

=10000(1+100C2+)

obviously 10000 is the correct choice.

Hence, the answer is the option (3).

Example 3: (1+x)nnx1 is divisible by (where nN ):

1) 2x

2) x2

3) 2x3

4) All of these

Solution:

(1+x)n=1+nx+n(n1)2!x2+n(n1)(n2)3!x3+

(1+x)nnx1

=x2[n(n1)2!+n(n1)(n3)3!x+.]

From above it is clear that (1+x)nnx1 is divisible by x2.

Trick: (1+x)nnx1. Put n=2 and x=3; Then 422.31=9 is not divisible by 6,54 but divisible by 9 . Which is given by option (b) i.e., x2=9

Hence, the answer is the option (2).

Example 4: 49n+16n1 is divisible by:

1) 3

2) 19

3) 64

4) 29


Solution:

49n+16n1=(1+48)n+16n1

1+nC1(48)+nC2(48)2++nCn(48)n+16n1

=(48n+16n)+nC2(48)2+nC3(48)3++nCn(48)n

=64n+82[nC262+nC3638+nC46482++nCn6n8n2]

Hence, 49n+16n1 is divisible by 64 .

Hence, the answer is option (3).

Example 5: The greatest integer m such that 5m divides 72n+23n33n1 for nN, is

1) 0

2) 1

3) 2

4) None of these

Solution:

72n+23n33n1=49n+24n1

=(501)n+(251)n1

= Multiple of 50+(1)n+ Multiple of 25+(1)n1

= a multiple of 25, since (1)n+(1)n1=0

Hence, the answer is option (1).


Articles

Get answers from students and experts
Back to top