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 ra^sb^t, 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  ⁿC_rx^n-ry^r

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

• ⁿC_r(-1)^r x^n-r. y^r

• ⁿC_r x^r

•   (-1)^{r n}C_r x^r

• (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 T_m 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: T_r+1 term is said to be numerically greatest for a given value of x, y provided T_r+1 ≥ T_r and T_r+1 ≥ T_r+2

i.e.

T_r+1/T_r = ⁿC_rx^n-ry^r/ⁿC_r-1(x)^n-r+1y^r-1 = (n-r+1)/r |(y/x)|

Properties of ⁿC_r:

• ⁿC_r = ⁿC_n-r menas ⁿC_x = ⁿC_y has two solutions x = y or x + y = n.

• ⁿC_r + ⁿC_r-1 = ⁿ⁺¹C_r

• ⁿC_r = n/r (^n-1C_r-1)

• ⁿC_r/ⁿC_r-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 = ⁿC₄a^n-4(-b)⁴

The sum of the 6th term is given by

T₆ = ⁿC₅a^n-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 C_n = 0, n > 2, where C_r = ⁿC_r.

Solution: We know that by the binomial theorem

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

Multiplying it by x we get,

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

Differentiating both sides we get,

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

Again multiplying by x we get,

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

Putting x = -1, we get

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

 

Illustration 4: If (1+x)ⁿ = C₀ + C₁x + C₂x² + C₃x³ + …. + C_nxⁿ, then prove that C₁² + 2C₂² + 3C₃² + ….….…... + nC_n² = (2n-1)!/[(n-1)!]²

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

Differentiating both sides with respect to x, we get

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

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

and (x+1)ⁿ = C₀xⁿ + C₁x^n-1 + C₂x^n-2 + C₁x^n-1 + ….… + C_n                   ….….(2)

Multiplying (1) and (2), we have

n(1 + x)^2n-1 = (C₁ + 2C₂x + 3C₃x² + …. + nC_nx^n-1) x (C₀xⁿ + C₁x^n-1 + C₂x^n-2 + C₃x^n-3 + ….. + C_n)  ….. (3)

Now, coefficient of x^n-1 on R.H.S.

C₁² + 2C₂² + 3C₃² + ….. + nC_n²

and coefficient of x^n-1 on L.H.S = n.^2n-1C_n-1

 

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

So, C₁² + 2C₂² + 3C₃² + …. + nC_n² = (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. ⁿC_r + ⁿC_r-1 =

(a) ⁿCr-1

(b) ⁿ⁺¹C_r

(c) ⁿ⁺¹C_r+1

(d) ⁿ⁺¹C_r-1

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

(a) T_r+1 = ⁿC_rx^n-r.y^r

(b) T_r+1 = ⁿC_r-1x^n-r.y^r

(c) T_r+1 = ⁿC_rx^n-r.y^r

(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.

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