JEE Main Study Notes for Binomial Theorem include binomial expansion, binomial coefficients, and binomial series. The topic Binomial Theorem is easier in comparison to the other chapters under Algebra. Binomial Theorem is a speedy method of growing a binomial expression with huge powers.

• The topic Binomial Theorem has a weightage of 1-2% in JEE Main.
• In JEE Main Mathematics Question Paper, Binomial Theorem carries a total of around 4 marks from Binomial Theorem.

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What is a Binomial and a Binomial Expression?

A special case of "polynomial" is the word "binomial" Polynomial implies an algebraic expression that consists of two or more algebraic terms. The binomial is also known as an algebraic expression with two distinct terms.

• Therefore, a binomial expression can be defined as an algebraic expression that consists of two distinct terms.
• In this definition, the phrase “distinct terms” is very essential to note.
• The phrase here means that either the two terms must have different variables or different powers on their variables.

For example,

x + y, x2 + y3 and x + x2 are expressions that have two distinct terms. Thus, they are considered to be binomial expressions. On the other hand, x + 2x or x2y + 2 x2y can be merged into one term. Thus, they are not examples of binomial expressions.

What is Binomial Theorem?

In algebra, the binomial theorem concentrates on the expansion of exponents or powers in a binomial expression. This theorem was provided by Newton. He explained the expansion of (x+y)n for distinct values of n.

According to his theorem, the general term in the expansion of (x+y)n could be represented in the form of pxq yr, where q and r are the non-negative integers. It also satisfies that q + r = n. Here, ‘p’ is known as the binomial coefficient.

(a+b)n

In (a+b)n , a + b is binomial. Thus, the expansion of (a+b)n could be quickly found using the binomial theorem.

It is important to understand how the formula of binomial expansion was derived in order to be able to solve questions with more ease.

On close examination of the expansion of (a + b) for distinct exponents, it is seen that,

For (a+b)0 = 1

For (a+b)1 = a + b

For (a+b)2 = a2 + 2ab + b2

For (a+b)3 = a3 + 3a2b + 3ab2 + b3

For (a+b)4 = a4 + 4a3b + 6a2b2+ 4ab3+ b4

From the expansion given above, some significant properties can be noted. These are -

• The number of terms in the expansion of (a+b)n is n + 1, that is if n = 3, the number of terms will be 3 + 1 = 4 and the process goes on like this.
• The power of a begins from n and reduces till it turns 0.
• In a similar manner, the power of b begins from 0 and rises till n.
• The binomial coefficients of the terms equidistant from the starting and the end are equal. For example, in (a+b)4 the binomial coefficients of a4 and b4,a3b, and ab3 are equal.
• The sum of the powers of its variables on any term is equal to n.

The triangle given above is known as Pascal’s Triangle. This triangle expresses the binomial coefficients for the expansion of distinct powers of (a + b).

The first row of the triangle shows the expansion of the binomial theorem for the power zero. In a similar manner, the second row shows the binomial coefficient of the terms for the power one.

From the figure given above, it is evident that any number in the triangle is an addition to the two numbers above it.

Before you generalize the formula for binomial expansion, note that the binomial coefficients are the values of nCr for distinct values of r.

Now, the binomial theorem can be generalized for any non-negative power n.

(x+y)n = nC0xn + nC1xn-1y + nC2xn-2y2 + … + nCrxn-ryr+ … + nCnxn-n yn.

From the equation given above,

First term, T1= nC0xn

Second term, T2 = nC1xn-1y1

Third term, T3= nC2xn-2y2

Therefore, the general term in the expansion of (x+y)n is,

Tr+1= nCrxn-ryr

Number of terms the expansion of (x+y+z+w)10

The problem given to understand these types of questions is an expansion of a quadrinomial or a polynomial that has four terms.

Going back to a property of the binomial theorem, the sum of the powers of its variables on any term is equal to n, where n is the exponent on (x + y).

Thus, in the expansion of (x+y)n,

(x+y)n = nC0xn + nC1xn-1y + nC2xn-2y2+ … + nC2xn-ryr+ … + nCnxn-nyn .

It is seen that the power n is allocated to the variables x and y in each permutation. Thus, this distribution can be associated with the coin–beggar’s method of permutation and combination.

According to the coin–beggar’s method, the number of ways to distribute n identical coins to p beggar’s is n+p-1Cp-1.

In this case, the number of coins is the exponent of any polynomial and the number of beggars is the number of terms in the polynomial.

Thus, for (x+y+z+w)10 , n = 10 and p = 4.

Therefore, the number of terms in its expansion would be = 10+4-1C4-1 = 13C3

This method could even be verified for the binomial expansion, where we have an exponent and p = 2. Thus, the number of terms = n+1C1= n+1 terms.

Must Read:

• JEE Main Study Notes for Algebra
• JEE Main Study Notes for Sequence & Series

Binomial Coefficient

In (a+b)2 = 1a2+ 2ab + 1b2

1, 2 and 1 are known as the binomial coefficients of a2, ab and b2 respectively.

Likewise, in (a+b)3 = 1a3+ 3a2b + 3ab2 + 1b3

1, 3, 3 and 1 are known as the binomial coefficients of a3, a2b, ab2 and b3 respectively.

Generally, in the expansion of (x+y)n

(x+y)n = nC0xn + nC1xn-1 y + nC2xn-2y2 + … + nCrxn-r yr + … + nCnxn-nyn .

all of the terms contain a constant that is multiplied with the variables in the form of nCr. These nCr variables are known as the binomial coefficients of distinct terms (according to the value of r).

Another essential point to note is that the sum of the binomial coefficients could be easily calculated through a replacement of the variables to 1.

For example,

The sum of the binomial coefficients of (x+y)n can be calculated in the following manner -

Write x = 1 and y = 1 in the expansion of (x+y)n , and you will get,

(1+1)n = 2n = nC0 + nC1 + nC2 + … + nCr + … + nCn

Check:

• JEE Main Study notes for Integral Calculus
• JEE Main Study Notes on Differential Calculus

How to use Pascal's triangle?

As studied before, Pascal’s triangle is a triangular allotment of binomial coefficients for the expansion of distinct powers.

Given below is Pascal's triangle for expansion until the exponent five.

Note that the triangle can be expanded for any number of rows. Every row expresses the binomial coefficient for any particular exponent.

For example, the first and second rows express the binomial coefficients for the expansion of (x + y) having the exponent 0 and 1 respectively.

Binomial Series

The binomial series is a special series in mathematics. It is also known as the Maclaurin series.

This series is a unique case of the binomial theorem, where x = 1 and y = x

Binomial Series:

(1+x)n = nC0 + nC1x + nC2x2 + …… nCrxr+ …. + nCnxn

According to the values of x and n, the series could be either converging or diverging.

Also Check: JEE Main Scalers and Vectors Study Notes

Applying Binomial Theorem if n is Negative or Fractional

The Binomial Theorem for the expansion of (x+y)n where n ∉ I+ can be expanded as,

The general term here is,

Note: Here, the binomial coefficients cannot be found using nCr directly, because this is not defined for the negative n.

For (a+x)n where n ∉ I+,

Here, where n is a non-positive integer, the series will converge only when (x/a) <1. At the same time, the number of terms will be infinite, that is, an infinite series.

For (1+x)n , where n ∉ I+,

The series given above converges for |x| < 1.

From the given series, a few important series can be found -

• (1+x)-1 = 1 – x + x2 – x3+ x4 - ……..∞
• (1-x)-1 = 1 + x + x2+ x3 + x4 - ……..∞
• (1+x)-2 = 1 – 2x + 3x2– 4x3 + 5x4 - ……..∞
• (1-x)-2 = 1 + 2x + 3x2 + 4x3+ 5x4 - ……..∞

Finding the Term Independent of x

The term independent of ‘x’ means the term in the binomial expansion that does not have any variable x included in it.

For example,

(x+y)2= x2 + 2xy + y2, the third term, that is, y2 is the term that is independent of ‘x’. The first term, that is, x2 is the term independent of y.

For example,

(x+y)3 = x3 + 3x2y + 3xy2 + y3.

As given above, the first term is independent of y and the fourth term is independent of x. The second and fourth terms include both the variables x and y. Thus, they are not independent of either x or y.

To find the independent term in a binomial expression of any power, follow the steps given below -

• For the binomial given with any power, note down its general term.
• Join all the ‘x’ terms with the use of the laws of exponent (if not combined).
• Equate the power or indices of ‘x’ to zero in order to find the value of the variable ‘r’.
• Once you know r, if r is a positive integer, then, (Tr+1)th term will be the term that is independent of x. Otherwise, no such term will exist.

For example,

Which term is independent of x in the expansion of (x – x2)10?

Step 1: Write its general term,

Tr+1 = 10Crx10.-r (-1)r(x2)r

Step 2: Combine all the x terms,

Tr+1 = 10Crx10-r + 2r (-1)r = 10Crx10+ r (-1)r

Step 3: Equate the power of x to zero

10 + r = 0 that is, r = -10 which is not a positive integer.

Therefore, no such term exists.

This step-by-step method is quite helpful to also find any term or its coefficient that has the specific power of x.

For example, if you are asked to find the coefficient of x15, it can be solved as given below -

Given, Tr+1 = 10Crx10-r + 2r (-1)r = 10Crx10 + r (-1)r

For x15, it must be equated to the power of x to 15.

That is, 10 + r = 15.

Thus, r = 5, that is, 6th term.

Therefore, the coefficient = T6 = -10C5

Points to Remember for Binomial Theorem

• When x, y ∈ R and n ∈ N, the binomial theorem is given as,

(x+y)n = nCrxn+ nC1xn-1y+ nC2xn-2y2 +…… … .. + nCrxn-ryr+ ….. + nCnyn

This could be written as ΣnCrxn-ryr. This is also known as the binomial theorem formula can be used for solving many problems.

• A few primary properties of the binomial expansion of the term (x+y)n are -

1. The number of terms in the expansion is (n+1), that is, it is one more than the index.
2. The sum of indices of x and y is typically n.
3. The binomial coefficients of the terms that are equidistant from the start and end are always equal. The reason for 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).

• Such an expansion generally goes by a simple rule -

1. The subscript of C, that is, the lower suffix of C is typically equal to the index of y.
2. Index of x = n – (lower suffix of C).

• The (r +1)them in the expansion of the expression (x+y)n is known as the general term and is represented as Tr+1 = nCrxn-ryr
• The term independent of x is evident without x. It is that value of r for which the exponent of x is zero.
• The middle term of the binomial coefficient varies according to the value of n. Two different cases can arise depending on whether n is even or n is odd. These are -

1. When n is even, the total number of terms is odd. In this case, there is a single middle term which is (n/2 +1)th and is represented as nCn/2 an/2 xn/2.
2. When n is odd, the total number of terms is even and there are two middle terms [(n+1)/2]thand [(n+3)/2]th which are equal to nC(n-1)/2 a(n+1)/2 x(n-1)/2 and nC(n+1)/2 a(n-1)/2 x(n+1)/2

• The binomial coefficient of the middle term is the highest binomial coefficient of the expansion.
• A few of the standard binomial theorem formulas that must be memorized are given below -

1. C0+ C1+ C2+ ….. + Cn= 2n
2. C0+ C2 + C4 + ….. = C1+ C3+ C5+ ……….= 2n-1
3. C02+ C12+ C22+ ….. + Cn2 = 2nCn= (2n!)/ n!n!
4. C0Cr+ C1Cr+1+ C2Cr+2+ ….. + Cn-rCn=(2n!)/ (n+r)!(n-r)!
5. One other result that can be applied to binomial theorem problems is nCr+ nCr-1 = n+1Cr
6. mC0 can also be replaced by m+1C0m+1C0 because the numerical value of both is the same, that is, 1. Likewise, mCm can be replaced by m+1Cm+1.
7. Note that (2n!) = 2n. n! [1.3.5. … (2n-1)]

• To be able to compute numerically the highest term in a binomial expansion of (1+x)n, find Tr+1 / Tr= (n – r + 1)x/r. Then, write the absolute value of x and find the value of r that is consistent with the inequality Tr+1/ Tr> 1.
• If the index n is something apart from a positive integer like a negative integer or fraction, then the number of terms in the expansion of (1+x)n becomes infinite.
• The expansions in ascending powers of x hold valid only when x is small. If x is huge, that is, |x| > 1 it becomes easy to expand in powers of 1/x which is small.
• The binomial expansion for the nth degree polynomial is expressed as –

• The expansions given below must be remembered for |x| < 1 -

1. (1+x)-1 = 1 – x + x2– x3+ x4- ….. ∞
2. (1-x)-1 = 1 + x + x2+ x3+ x4+ ….. ∞
3. (1+x)-2 = 1 – 2x + 3x2– 4x3+ 5x4 - ….. ∞
4. (1-x)-2 = 1 +2x + 3x2+4x3+ 5x4+ ….. ∞

Check: JEE Main Probability Study Notes

Solved Examples on Binomial Theorem

Question 1: Calculate the coefficient of the independent term of x in the expansion of (3x - (2/x2))15.

Solution: The general term of (3x - (2/x2))15 is Tr+1 = 15Cr (3x)15-r (-2/ x2)r. It is independent of x when,

15 - r - 2r = 0 => r = 5

Therefore, T6 =15C5(3)10(-2)5 = -16C5 310 25.

Question 2: Suppose the coefficients of (2r + 4)th and (r - 2)th terms in the expansion of (1+x)18 are equal, find the value of r.

Solution: The general term of (1 + x)n is Tr+1= Crxr

Thus, the coefficient of (2r + 4)th term is,

T2r+4 =T2r+3+1 = 18C2r+3

and the coefficient or (r - 2)th term is,

Tr-2 = Tr-3+1 = 18Cr-3.

=> 18C2r +3 =18Cr -3.

=> (2r + 3) + (r-3) = 18 (·.· nCr = nCK => r = k or r + k = n)

Therefore, r = 6

Question 3: Suppose a1, a2, a3, and a4 are the coefficients of any four consecutive terms in the expansion of (1+x)n, prove that - a1/(a1+a2) + a2/(a3+a4) = 2a2/(a2+a3)

Solution: Because a1, a2, a3 and a4 are the coefficients of consecutive terms,

Let a1= nCr

a2 = nCr+1

a3 nCr+2 and

a4 = nCr+3

a1/(a1+a2) = nCr/(nCr+nCr+1) = 1/(1+((n-r)/(r+1))) = (r+1)/(n+1)

Likewise, a2/(a2+a3) = (r+3)/(n+1)

Now, a3/(a3+a4) + a1/(a1+a2) = (2r+4)/(n+1)

= 2(r+1)/(n+1) = 2a2/(a2+a3)

Hence, proved.

Question 4: Find out which one among them is larger - 9950 + 10050 or 10150.

Solution: Find out 10150 - 9950 in terms of remaining term, that is,

10150 - 9950 = (100+1)50 - (100 - 1)50

= (C0.10050 + C110049 + C2.10048 +......)

= (C010050 - C110049 + C210048 -......)

= 2[C1.10049 + C310047 +.........]

= 2[50.10049 + C310047 +.........]

= 10050 + 2[C310047 +............]

> 10050

=> 10150 > 9950 + 10050

Question 5: Find the value of the greatest term in the expansion of √3(1+(1/√3))20.

Solution: Assume that Tr+1 is the greatest term, then Tr < Tr+1 > Tr+2

Consider : Tr+1 > Tr

=> 20Cr (1/√3)r > 20Cr-1(1/√3)r-1

=> ((20)!/(20-r)!r!) (1/(√3)r) > ((20)!/(21-r)!(r-1)!) (1/(√3)r-1)

=> r < 21/(√3+1)

=> r < 7.686 ......... (i)

Likewise, consider Tr+1 > Tr+2

=> r > 6.69 .......... (ii)

From (i) and (ii),

r = 7

Hence, the greatest term = T8 = 25840/9

Tricks to Solve Questions from Binomial Theorem

Question 1: Prove that  (-3)r-1 3nC2r-1= where k = 3n/2 and n is an even positive integer.

The trick to solve this question: Since it is given that n is an even integer, assume that n = 2m. Substitute this value in the equation given for k. Find the summation (S). After you find two expressions from (1+x)6m and (1-x)6m, label them as (i) and (ii). Then consider that x2 = y. You will find that y = -3. Then, RHS = S. Prove with what you have that LHS = RHS.

Solution: As n is an even integer, let n = 2m,

k = 3n/2 = 6m/2 =3m

The summation is,

S =

Now, (1+x)6m = 6mC0 + 6mC1 x + 6mC2 x2 +...

...+ 6mC6m-1 x6m-1 + 6mC6m x6m ...... (i)

(1-x)6m = 6mC0 + 6mC1 (-x) + 6mC2 (-x)2 +...

...+ 6mC6m-1 (-x)6m-1 + 6mC6m (-x)6m ...... (ii)

=> (1+x)6m - (1-x)6m = 2[6mC1x + 6mC3x2 +...+ 6mC6m-1x6m-2]

((1+x)6m - (1-x)6m)/2x = 6mC1 + 6mC3 x2 +......+ 6mC6m-1 x6m - 2

Let x2 = y

=>((1 + √y)6m-(1-√y)6m)/2√y=6mC1+6mC3y + 6mC3 y+......6mC6m-1 y3m-1

With y = -3, RHS = S.

LHS = 

Hence, proved.

Question 2: Which term is independent of x in the expansion of (x – 1/x)20?

Trick to solve this question: First, write down the general term of the given expansion. Then, combine all the x terms. Equate the power of x to zero and you will find the term.

Solution: r+1 = 11th term is independent of x.

Tips for JEE Main Mathematics Preparation

1. Master your concepts in Calculus and Algebra because they have the highest weightage in the Mathematics section and are also easy to score.
2. Maintain a formula book where you have noted down all the formulas that you have come across while studying. Revise them every day.
3. Refer to R.D Sharma, Arihant, and NCERT for preparation.
4. Attempt mock tests and previous years’ question papers often.

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