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Binomial Theorem

Binomial Theorem is one of the most important chapters of Algebra in the syllabus of IIT JEE. The beginners sometimes find it difficult as this topic is very new to them. The other reason is that the concepts of Combinations are also used in this theorem. This chapter can be said to be one of the easiest as it has very few twists and turns.

The chapter begins with the Introduction to Binomial Theorem which is followed by the Properties of the Binomial Theorem. Binomial Coefficients are the most important topic of Binomial theorem. The sum of the Binomial coefficients, the coefficient of a particular term, greatest Binomial coefficient etc. has been discussed at length. However, some important results have been given at the end which is supplemented by the application of Binomial Expression. Solved examples as usual are very useful as they can reappear in the examination with slight modification and they also facilitate learning.

View the following video for more on Binomial Theorem

What is a Binomial Expression?

An algebraic expression consisting of two different terms is called a binomial expression. Eg: Terms of the form x3 + y3 are binomial expressions.

What is the Binomial Theorem for a positive integral?

The binomial theorem explains the way of expressing and evaluating the powers of a binomial. This theorem explains that a term of the form (a+b)n can be expanded and expressed in the form of rasbt, where the exponents s and t are non-negative integers satisfying the condition s + t = n. The coefficient r is a positive integer. The terms involved are called binomial coefficients and since it is for positive indices it expands only the positive powers.

The general binomial expansion for any index is given by

(x+y)n = nC0xny0 + nC1x(n-1)y1 + nC2x(n-2)y2 + …….. + nC(n-1) x1y(n-1) + nCnx0yn.

Binomial Theorem is a quite vast chapter and it has been divided into several sections like:

Introduction To Binomial Theorem

Properties Of Binomial Expansion

Binomial Coefficients

Sum Of Binomial Coefficients

Coefficient Of A Particular Term

Middle Term

Greatest Binomial Coefficient

Application Of Binomial Expression

Solved Examples

We shall be discussing these topics here in brief as they have been discussed in detail in the coming sections.

Important Points of Binomial Expansion of the term (x + y)n:

Number of terms in the expansion of (x + y)n is (n + 1) i.e. one more than the index.

Sum of indices of x and y in each term in the expansion of (x + y)n is n.

The binomial coefficients of the terms which are equidistant from the starting and the end are always equal. The simple reason behind this is

C (n, r) = C (n, n-r) which gives C (n, n) C (n, 1) = C (n, n-1) C (n, 2) = C (n, n-2).

We have   nCrxn-ryr

In this, if we replace y by -y, we have

 nCr(-1)r xn-r. yr

 nCr xr

  (-1)r nCr xr

(1 + x)n + (1 - x)n = 2[nC0 + nC2x2 + …..  ]

(1 + x)n - (1 – x)n = 2[nC1 + nC3x3 + nC5x5 + …. …..  ]

Such an expansion always follows a simple rule which is

The subscript of C i.e. the lower suffix of C is always equal to the index of y and

Index  of x = n – (lower suffix of C).

Important Terms in Binomial:

General Term

Term independent of x

Middle term

(a) General Term: T(r+1)th term is called as general term in (x + y)n and the general term is given by

T(r+1) = nCrxn-ryr

(b) Term independent of x: It means the term containing x0.

(c) Middle Term: Let Tm be the middle term in expansion of (x+y)n, then

Case 1: If n is odd, then the number of terms will be even and so there will be two middle terms given by [(n+1)/2] th and [(n+3)/2] th.

Case 2: If n is even, then the number of terms will be odd and so there will be only one middle term i.e. (n/2 + 1)th.

Note: The binomial coefficient of the middle term is the greatest binomial coefficient of the expansion.

Numerically greatest Term in (x+y)n: Tr+1 term is said to be numerically greatest for a given value of x, y provided Tr+1 ≥ Tr and Tr+1 ≥ Tr+2

i.e. 

Tr+1/Tr = nCrxn-ryr/nCr-1(x)n-r+1yr-1 = (n-r+1)/r |(y/x)|

Properties of nCr:

nCr = nCn-r menas nCx = nCy has two solutions x = y or x + y = n.

nCr + nCr-1 = n+1Cr

nCr = n/r (n-1Cr-1)

nCr/nCr-1 = (n-r+1)/r

How to find the nature of integral part of expression N = (a + √b)n  (n ∈ N):

In order to find the integral part of the above expression, follow the below listed steps:

Step 1: Consider N’ = (a-√b)n or (√b-a)n according as a > √b or √b > a.

Step 2: Use N + N’ or N – N’ such that the result is an integer.

Step 3: Use the fact N = I + f, where I stands for [N] and ‘f’ for {N}.

Let us have a look at some of questions that can be asked from this section:

Illustration 1: (a) Find the coefficient of x201 in the expansion of 1 + (1 + x) + (1 + x)2 + ….. (1 + x)2011.

(b) Find the coefficient of x301 in the expansion of (1 + x) + 2(1 + x)2 + 3(1 + x)3 + ….. … + 500 (1 + x)500.

Solution: 1 + (1 + x) + (1 + x)2 + ….. (1 + x)2011





So, the coefficient of x201 in 1/x . (1 + x)2012 – 1/x

⇒ The coefficient of x202 in (1 + x)2012  ⇒ 2012C202.

(b) S = (1 + x) + 2(1 + x)2 + 3(1 + x)3 + ….. … + 500 (1 + x)500.                              ….….. (1)

(1 + x)S = (1 + x)2 + 2(1 + x)3 + ….. … + 499 (1 + x)500 + 500 (1 + x)501                 ….….. .(2)

Subtracting (2) from (1), we have

S(-x) = (1 + x) + (1 + x)2 + (1 + x)3 + ….. … + (1 + x)500 – 500(1 + x)501

= (1 + x) [{(1 + x)500 – 1}/ (1 + x – 1)] – 500(1 + x)501

S(-x) = (1 + x)[(1 + x)500 – 1]/ x – 500(1 + x)501

So, S = 1/x2 (1 + x)(1 – (1 + x)500) + 500. 1/x . (1 + x)501

= 1/x2 - (1 + x)500/x2 + 1/x – (1 + x)500/x + 500. 1/x (1 + x)501

So, coefficient of x301 = – 500C303 –  500C302 + 500. 501C302

= 500. 501C302 – (500C303 + 500C302)

= 500. 501C302 – 501C303.

Illustration 2: In the binomial expansion of (a-b)n, n ≥ 5, the sum of the 5th and 6th terms is zero. Then find the value of a/b.

Solution: The sum of the 5th term is given by

T5 = nC4an-4(-b)4

The sum of the 6th term is given by

T6 = nC5an-5(-b)5

It is given in the question that T5 + T6 = 0.

This gives a/b = (n-4)/5.

Illustration 3: Prove that C0 – 22C1 + 32C2 - ….. + (-1)n(n+1)2 Cn = 0, n > 2, where Cr = nCr.

Solution: We know that by the binomial theorem

(1 + x)n = C0 + C1x + C2x2 + …. + Cnxn

Multiplying it by x we get,

x(1 + x)n = C0x + C1x2 + C2x3 + ……. + Cnxn+1

Differentiating both sides we get,

(1 + x)n + nx(1 + x)n-1 = C0 + 2C1x + 3C2x2 + ….. + (n+1)Cnx

Again multiplying by x we get,

x(1+x)n + nx2(1+x)n-1 = C0x + 2C1x2 + 3C2x3 + ……. + (n+1)Cnxn+1

Hence, (1 + x)n + nx(1 + x)n-1 + 2 nx(1 + x)n-1 + n(n-1)x2(1 + x)n-2  = C0 + 22C1x + 32C2x2 + ….. + (n + 1)2 Cnxn

Putting x = -1, we get

0 = C0 – 22C1 + 32C2 - ….. + (-1)n(n+1)2Cn

Illustration 4: If (1+x)n = C0 + C1x + C2x2 + C3x3 + …. + Cnxn, then prove that C12 + 2C22 + 3C32 + ….….…... + nCn2 = (2n-1)!/[(n-1)!]2

Solution: Given (1+x)n = C0 + C1x + C2x2 + C3x3 + …. + Cnxn

Differentiating both sides with respect to x, we get

n(1+x)n-1 = 0 + C1 + 2C2x + 3C3x2 + …. + nCnxn-1

n(1+x)n-1 = C1 + 2C2x + 3C3x2 + …. + nCnxn-1                                    ….… (1)

and (x+1)n = C0xn + C1xn-1 + C2xn-2 + C1xn-1 + ….… + Cn                   ….….(2)

Multiplying (1) and (2), we have

n(1 + x)2n-1 = (C1 + 2C2x + 3C3x2 + …. + nCnxn-1) x (C0xn + C1xn-1 + C2xn-2 + C3xn-3 + ….. + Cn)  ….. (3)

Now, coefficient of xn-1 on R.H.S.

C12 + 2C22 + 3C32 + ….. + nCn2

and coefficient of xn-1 on L.H.S = n.2n-1Cn-1



But, we know that (3) is an identity and therefore the coefficient of xn-1 in R.H.S. = coefficient of xn-1 in L.H.S.

So, C12 + 2C22 + 3C32 + …. + nCn2 = (2n-1)!/{(n-1)!}.

Q1. The number of terms in the expansion of (x + y)n is

(a) one more than the index

(b) one less than the index

(c) same as index

(d) none of these

Q2. nCr + nCr-1 =

(a) nCr-1

(b) n+1Cr

(c) n+1Cr+1

(d) n+1Cr-1

Q3. The general term in (x + y)n is given by

(a) Tr+1 = nCrxn-r.yr

(b) Tr+1 = nCr-1xn-r.yr

(c) Tr+1 = nCrxn-r.yr

(d) none of these

Q4. The middle term in expansion of (x + y)n, in case of n being odd is

(a) (n+3)/2 th

(b) n/2 th

(c) (n/2 + 1) th

(d) (n+1)/2 th and (n+3)/2 th

Q5. The middle term in expansion of (x + y)n, in case of n being even is

(a) (n/2 + 1) th

(b) n/2 th

(c) (n+1)/2 th

(d) none of these

Q1.
Q2.
Q3.
Q4.
Q5.

(a)
(b)
(c)
(d)
(a)

Binomial Theorem is important from the perspective of scoring high in IIT JEE as there are few fixed pattern on which a number Multiple Choice Questions are framed on this topic. The chapters of Binomial Theorem, Permutations and Combinations, and Probability together fetch 10-20 marks varying from one examination to the other.

Related resources:

Click here for the Entire Syllabus of IIT JEE Mathematics.

Look into the Past Year Papers with Solutions to get a hint of the kinds of questions asked in the exam.

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