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Mathematics Grade XI

Grade 11 Mathematics

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Chapter 7: Binomial Theorem

Class 11 • Mathematics • Chapter 7

Binomial Theorem

A fast way to expand a bracket raised to a power, without multiplying it out by hand.

CBSE / NCERT Aligned   School Revise

Chapter Roadmap

Expanding Brackets • Pascal’s Triangle • Binomial Coefficients • General Term • Middle Term • Term Independent of x • Key Results • Applications
1

Why the Binomial Theorem Matters

You already know that (a + b)² = a² + 2ab + b². Multiplying out (a + b)³ by hand is still manageable, but (a + b)⁷ or (a + b)¹⁰ would take a very long time and you would almost certainly make a slip somewhere. The Binomial Theorem gives a single formula that writes down the whole expansion at once, no matter how large the power, as long as it is a positive whole number.

The pattern behind the expansion is beautiful and orderly. The powers of the first term step down while the powers of the second term step up, and the numbers in front, called the binomial coefficients, follow the rows of Pascal’s triangle. Once you see this pattern you can pick out any single term you want, such as the middle term or the term that has no x in it, without writing the full expansion. That skill is used constantly in board exams and in competitive papers like JEE.

Key idea
For a positive whole number n, (a + b)ⁿ expands into (n + 1) terms. In each term the power of a falls from n to 0 while the power of b rises from 0 to n, and the two powers always add up to n. The number in front of each term is a binomial coefficient ⁿCᵣ.
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Key Terms You Must Know
Term Meaning Example
Binomial An expression with exactly two terms. a + b, or 2x − 3
Binomial Theorem The rule that expands (a + b)ⁿ for a positive whole number n. (a + b)² = a² + 2ab + b²
Binomial coefficient The number ⁿCᵣ in front of a term; it equals n!/(r!(n − r)!). ₆C₂ = 15
Pascal’s triangle A triangle of numbers whose rows give the binomial coefficients. Row 3 is 1, 3, 3, 1
General term The formula for the (r + 1)th term, Tₛ₊₁ = ⁿCᵣ aⁿ₋ᵣ bᵣ. Lets you find any single term
Middle term The central term (or two central terms) of the expansion. Depends on whether n is even or odd
Independent of x A term that contains no x, that is, x to the power 0. The constant term in (x + 1/x)₆
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Core Concepts, Step by Step

1. Expanding a Bracket Raised to a Power

The Binomial Theorem states that for any positive whole number n, (a + b)ⁿ = ⁿC₀ aⁿ + ⁿC₁ aⁿ₋¹ b + ⁿC₂ aⁿ₋² b² + … + ⁿCₙ bⁿ. Look at the structure: the power of a starts at n and drops by one each time, the power of b starts at 0 and rises by one each time, and in every single term the two powers add up to n. The coefficients ⁿC₀, ⁿC₁, ⁿC₂ and so on are the binomial coefficients. There are always (n + 1) terms.

2. Pascal’s Triangle

Before reaching for the formula, there is a lovely shortcut for the coefficients. Write 1 at the top, then each new row starts and ends with 1, and every number inside is the sum of the two numbers just above it. The row labelled n gives exactly the coefficients of (a + b)ⁿ. For instance, the row 1, 3, 3, 1 gives (a + b)³ = a³ + 3a²b + 3ab² + b³. The triangle below shows the rows from n = 0 to n = 5.

Pascal’s triangle: rows give the binomial coefficients

1 (n = 0)

1    1 (n = 1)

1    2    1 (n = 2)

1    3    3    1 (n = 3)

1    4    6    4    1 (n = 4)

1    5    10    10    5    1 (n = 5)

Each number is the sum of the two numbers just above it (for example 2 = 1 + 1, and 6 = 3 + 3). Row n gives the coefficients of (a + b) raised to the power n. So (a + b)³ uses the row 1, 3, 3, 1, giving a³ + 3a²b + 3ab² + b³.

3. Binomial Coefficients

Pascal’s triangle is quick for small powers, but for larger ones we use the formula directly. The coefficient of the term involving bᵣ is ⁿCᵣ = n!/(r!(n − r)!), the same combination you met in the previous chapter. This makes sense: choosing which r of the n brackets contribute a b is exactly a combination. Because ⁿCᵣ = ⁿCₙ₋ᵣ, the coefficients read the same forwards and backwards, which is why every row of Pascal’s triangle is a mirror image of itself.

4. The General Term

The real power of the theorem is being able to write down any one term without the rest. The general term, which is the (r + 1)th term, is Tₛ₊₁ = ⁿCᵣ aⁿ₋ᵣ bᵣ. By choosing the right value of r you can reach into the expansion and pull out exactly the term you need. Notice that r starts at 0 (giving the first term), so the 4th term comes from r = 3, the 5th term from r = 4, and so on. Always set r one less than the term number you want.

5. The Middle Term

Since there are (n + 1) terms, where the middle sits depends on whether n is even or odd. If n is even, the number of terms is odd, so there is exactly one middle term, the (n/2 + 1)th. If n is odd, the number of terms is even, so there are two middle terms, the ((n + 1)/2)th and the ((n + 3)/2)th. To find a middle term you simply work out its position, set r one less, and use the general term.

6. The Term Independent of x

A common and important question asks for the term that has no x in it, called the term independent of x, or the constant term. The method is always the same: write the general term, gather all the powers of x into a single power, set that power equal to 0, and solve for r. That value of r, put back into the general term, gives the constant term. For example, in (x + 1/x)₆ the general term simplifies to ₆Cᵣ x⁶₋²ᵣ, which has no x when 6 − 2r = 0, that is r = 3.
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Key Results & Proofs

Three results are worth understanding rather than just learning. Each one follows quickly once the structure of the expansion is clear.

Result 1: The General Term Formula

Statement. In the expansion of (a + b)ⁿ, the (r + 1)th term is Tₛ₊₁ = ⁿCᵣ aⁿ₋ᵣ bᵣ.

Proof

Think about how a term containing exactly r factors of b is formed.

Writing (a + b)ⁿ means multiplying n identical brackets (a + b) together.  –  n copies of the same bracket
A term with bᵣ is made by picking b from r of the brackets and a from the other (n − r).  –  each bracket gives either a or b
The number of ways to choose those r brackets is ⁿCᵣ.  –  this is a combination
Tₛ₊₁ = ⁿCᵣ aⁿ₋ᵣ bᵣ coefficient times the two powers

The powers always satisfy (n − r) + r = n, a quick check that a term is written correctly.
Result 2: The Sum of the Coefficients

Statement. In the expansion of (1 + x)ⁿ, the sum of all the binomial coefficients is 2ⁿ.

Proof

Substituting a well-chosen value of x collapses the whole expansion.

(1 + x)ⁿ = ⁿC₀ + ⁿC₁x + ⁿC₂x² + … + ⁿCₙxⁿ binomial expansion
Put x = 1, since that turns every power of x into 1.  –  a clever substitution
(1 + 1)ⁿ = ⁿC₀ + ⁿC₁ + ⁿC₂ + … + ⁿCₙ left side becomes 2ⁿ
2ⁿ = sum of all the coefficients the result

Putting x = −1 instead shows that the alternating sum ⁿC₀ − ⁿC₁ + ⁿC₂ − … equals 0 when n ≥ 1.
Result 3: Pascal’s Rule

Statement. For whole numbers, ⁿCᵣ₋₁ + ⁿCᵣ = ⁿ₊¹Cᵣ.

Proof

This rule explains why every inside number of the triangle is a sum of the two above it.

This is exactly the rule that builds Pascal’s triangle.  –  each entry is the sum of the two above
Add the two combinations over a common denominator and simplify.  –  standard algebra of factorials
ⁿCᵣ₋₁ + ⁿCᵣ = n!/((r−1)!(n−r+1)!) + n!/(r!(n−r)!) Write each as a factorial expression
ⁿCᵣ₋₁ + ⁿCᵣ = (n+1)!/(r!(n+1−r)!) combine and simplify
ⁿCᵣ₋₁ + ⁿCᵣ = ⁿ₊¹Cᵣ this is the definition of (n+1)C r

It lets you build any row of Pascal’s triangle from the row before, without using factorials at all.
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Worked Examples
Example 1

Question: Expand (x + y)⁴ using the Binomial Theorem.

▶ Show full working

The coefficients come from row 4 of Pascal’s triangle: 1, 4, 6, 4, 1. The power of x falls from 4 to 0 while the power of y rises from 0 to 4.

Coefficients (row 4): 1, 4, 6, 4, 1.  –  from Pascal’s triangle
(x + y)^4 = x^4 + 4x^3 y + 6x^2 y^2 + 4x y^3 + y^4 apply the pattern
Check: in every term the powers add to 4.  –  quick verification

Answer: x⁴ + 4x³y + 6x²y² + 4xy³ + y⁴.
Example 2

Question: Expand (2x + 3)³.

▶ Show full working

Use row 3 coefficients 1, 3, 3, 1, taking a = 2x and b = 3. Be careful to raise the whole of 2x and of 3 to each power.

Coefficients (row 3): 1, 3, 3, 1.  –  from Pascal’s triangle
(2x + 3)^3 = (2x)^3 + 3(2x)^2(3) + 3(2x)(3)^2 + (3)^3 apply the pattern
  = 8x^3 + 3(4x^2)(3) + 3(2x)(9) + 27 work out each power
  = 8x^3 + 36x^2 + 54x + 27 simplify each term

Answer: 8x³ + 36x² + 54x + 27.
Example 3

Question: Expand (x − 2)⁴.

▶ Show full working

Take b = −2. The minus sign makes the terms alternate in sign, because even powers of −2 are positive and odd powers are negative.

Coefficients (row 4): 1, 4, 6, 4, 1, with b = −2.  –  note the minus sign
(x − 2)^4 = x^4 + 4x^3(−2) + 6x^2(4) + 4x(−8) + 16 apply the pattern
  = x^4 − 8x^3 + 24x^2 − 32x + 16 simplify

Answer: x⁴ − 8x³ + 24x² − 32x + 16.
Example 4

Question: Write the general term in the expansion of (x + 2)⁷.

▶ Show full working

Use Tₛ₊₁ = ⁿCᵣ aⁿ₋ᵣ bᵣ with n = 7, a = x and b = 2.

T(r+1) = 7Cr (x)^(7−r) (2)^r general term formula
T(r+1) = 7Cr 2^r x^(7−r) bring the constant to the front

Answer: Tₛ₊₁ = ₇Cᵣ · 2ᵣ · x⁷₋ᵣ.
Example 5

Question: Find the 4th term in the expansion of (x + 2)⁷.

▶ Show full working

The 4th term means r + 1 = 4, so r = 3. Substitute into the general term from Example 4.

r = 3 since the 4th term has r = 3
T4 = 7C3 (x)^(7−3) (2)^3 substitute r = 3
T4 = 35 × x^4 × 8 7C3 = 35 and 2^3 = 8
T4 = 280 x^4 multiply

Answer: 280x⁴.
Example 6

Question: Find the middle term in the expansion of (x + y)⁶.

▶ Show full working

Here n = 6 is even, so there is exactly one middle term, the (n/2 + 1)th = 4th term, which has r = 3.

n = 6 is even, so the middle term is the 4th, with r = 3.  –  one middle term
T4 = 6C3 (x)^(6−3) (y)^3 general term with r = 3
T4 = 20 x^3 y^3 6C3 = 20

Answer: 20x³y³.
Example 7

Question: Find the two middle terms in the expansion of (x + y)⁷.

▶ Show full working

Here n = 7 is odd, so there are two middle terms, the 4th and the 5th, coming from r = 3 and r = 4.

T4 = 7C3 (x)^4 (y)^3 r = 3 term
T4 = 35 x^4 y^3 7C3 = 35
T5 = 7C4 (x)^3 (y)^4 r = 4 term
T5 = 35 x^3 y^4 7C4 = 35

Answer: The two middle terms are 35x⁴y³ and 35x³y⁴.
Example 8

Question: Find the term independent of x in (x + 1/x)⁶.

▶ Show full working

Write the general term, collect the powers of x, and set the total power to 0.

T(r+1) = 6Cr (x)^(6−r) (1/x)^r general term
T(r+1) = 6Cr x^(6−r) x^(−r) since (1/x)^r = x^(−r)
T(r+1) = 6Cr x^(6−2r) add the powers of x
For no x, set 6 − 2r = 0, so r = 3.  –  power of x must be 0
term = 6C3 = 20 put r = 3

Answer: The term independent of x is 20.
Example 9

Question: Find the coefficient of x⁵ in the expansion of (x + 3)⁷.

▶ Show full working

The general term has x^(7 − r). Set 7 − r = 5 to land on x^5, then read off the coefficient.

T(r+1) = 7Cr (x)^(7−r) (3)^r general term
For x^5 we need 7 − r = 5, so r = 2.  –  match the power of x
coefficient = 7C2 × 3^2 put r = 2
coefficient = 21 × 9 7C2 = 21 and 3^2 = 9
coefficient = 189 multiply

Answer: 189.
Example 10

Question: Find the term independent of x in (x² + 1/x)⁹.

▶ Show full working

Collect the powers of x carefully, since the first term is x², then set the total power to 0.

T(r+1) = 9Cr (x^2)^(9−r) (1/x)^r general term
T(r+1) = 9Cr x^(18−2r) x^(−r) expand the powers
T(r+1) = 9Cr x^(18−3r) add the powers of x
For no x, set 18 − 3r = 0, so r = 6.  –  power of x must be 0
term = 9C6 = 9C3 = 84 put r = 6, using symmetry

Answer: The term independent of x is 84.
Example 11

Question: Evaluate (102)³ using the Binomial Theorem.

▶ Show full working

Write 102 as 100 + 2 and expand the cube. The arithmetic becomes easy because powers of 100 are simple.

(102)^3 = (100 + 2)^3 split into a convenient sum
  = 100^3 + 3(100)^2(2) + 3(100)(2)^2 + 2^3 binomial expansion of a cube
  = 1000000 + 60000 + 1200 + 8 work out each term
  = 1061208 add

Answer: 1061208.
Example 12

Question: Find the sum of all the coefficients in the expansion of (1 + x)₈.

▶ Show full working

Substitute x = 1, which turns every term into just its coefficient.

(1 + x)^8 = sum of coefficients when x = 1 by Result 2
(1 + 1)^8 = 2^8 put x = 1
2^8 = 256 evaluate

Answer: 256.
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Where You Meet This in Real Life

Quick mental estimates

Numbers like (1.02)⁵ or (99)³ can be worked out fast using the first few binomial terms, a trick used in finance for compound growth and in physics for small corrections.

Probability

The same binomial coefficients count the outcomes in coin tosses and similar experiments, which leads directly to the binomial probability distribution you will study later.

Computer science

Binomial coefficients count subsets and paths, and appear in algorithms, data compression and the analysis of how programs scale.

Algebra and calculus ahead

Expanding brackets neatly is the foundation for series, approximations and the binomial series for fractional powers that you meet in higher classes.

Pattern and design

Pascal’s triangle hides many patterns, including triangular numbers and the Fibonacci sequence, and is used in art and architecture for balanced, symmetric designs.
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Practice Sets A to D

Practice Set A – Basics

A1. Write the row of Pascal’s triangle for n = 4.

▶ Reveal full working

Start and end with 1; each inside number is the sum of the two above it in row 3 (1, 3, 3, 1).

Row 3 is 1, 3, 3, 1.  –  the row above
Add neighbours: 1, (1+3), (3+3), (3+1), 1.  –  Pascal’s rule
Row 4 is 1, 4, 6, 4, 1.  –  the answer

Answer: 1, 4, 6, 4, 1.

A2. Expand (a + b)³.

▶ Reveal full working

Use row 3 coefficients 1, 3, 3, 1.

(a + b)^3 = a^3 + 3a^2 b + 3a b^2 + b^3 apply the pattern

Answer: a³ + 3a²b + 3ab² + b³.

A3. Find the binomial coefficient ₆C₂.

▶ Reveal full working

Use ⁿCᵣ = n!/(r!(n − r)!).

6C2 = (6 × 5)/(2 × 1) combination formula
6C2 = 15 simplify

Answer: 15.

A4. Write the general term of (a + b)ⁿ.

▶ Reveal full working

This is the standard formula for the (r + 1)th term.

T(r+1) = nCr a^(n−r) b^r general term formula

Answer: Tₛ₊₁ = ⁿCᵣ aⁿ₋ᵣ bᵣ.

Practice Set B – Conceptual

B1. How many terms are there in the expansion of (a + b)ⁿ, and why?

▶ Reveal full working

Count the values that r can take.

The power r runs from 0 up to n.  –  r = 0, 1, 2, …, n
That is (n + 1) different values.  –  count them

Answer: There are (n + 1) terms.

B2. Why do the coefficients of (a + b)ⁿ read the same forwards and backwards?

▶ Reveal full working

Use the symmetry of binomial coefficients.

The coefficient of the term with b^r is nCr.  –  general term
By symmetry, nCr = nC(n − r).  –  symmetry rule
So the rth coefficient equals the coefficient counted from the other end.  –  mirror image

Answer: Because nCr = nC(n − r), so each row of Pascal’s triangle is symmetric.

B3. In (1 + x)ⁿ, what is the sum of all the coefficients, and how do you find it?

▶ Reveal full working

Substitute a convenient value of x.

Put x = 1 in the expansion.  –  clever substitution
(1 + 1)^n = 2^n every power of 1 is 1

Answer: The sum is 2ⁿ, found by putting x = 1.

B4. Where is the middle term when n is even, and when n is odd?

▶ Reveal full working

The number of terms is (n + 1).

If n is even, (n + 1) is odd, so there is one middle term, the (n/2 + 1)th.  –  odd number of terms
If n is odd, (n + 1) is even, so there are two middle terms.  –  even number of terms

Answer: Even n: one middle term, the (n/2 + 1)th. Odd n: two middle terms.

Practice Set C – Application / Numerical

C1. Find the 3rd term in the expansion of (x + 2)⁶.

▶ Reveal full working

The 3rd term has r = 2. Use the general term with n = 6.

T3 = 6C2 (x)^(6−2) (2)^2 general term, r = 2
T3 = 15 × x^4 × 4 6C2 = 15, 2^2 = 4
T3 = 60 x^4 multiply

Answer: 60x⁴.

C2. Find the middle term of (2x + 1)⁴.

▶ Reveal full working

n = 4 is even, so the middle term is the 3rd, with r = 2.

T3 = 4C2 (2x)^2 (1)^2 general term, r = 2
T3 = 6 × 4x^2 × 1 4C2 = 6, (2x)^2 = 4x^2
T3 = 24 x^2 multiply

Answer: 24x².

C3. Find the term independent of x in (x + 1/x)₈.

▶ Reveal full working

Collect the powers of x and set the total power to 0.

T(r+1) = 8Cr x^(8−r) x^(−r) general term
T(r+1) = 8Cr x^(8−2r) add the powers
Set 8 − 2r = 0, so r = 4.  –  power must be 0
term = 8C4 = 70 put r = 4

Answer: 70.

C4. Find the coefficient of x³ in the expansion of (x + 2)⁵.

▶ Reveal full working

The general term has x^(5 − r). Set 5 − r = 3.

T(r+1) = 5Cr (x)^(5−r) (2)^r general term
For x^3, set 5 − r = 3, so r = 2.  –  match the power
coefficient = 5C2 × 2^2 = 10 × 4 put r = 2
coefficient = 40 multiply

Answer: 40.

Practice Set D – HOTS / Word Problems

D1. Find the term independent of x in (2x² − 1/x)⁶.

▶ Reveal full working

Carefully collect the powers of x, remembering the minus sign and the factor 2.

T(r+1) = 6Cr (2x^2)^(6−r) (−1/x)^r general term
T(r+1) = 6Cr 2^(6−r) (−1)^r x^(12−3r) gather powers of x
Set 12 − 3r = 0, so r = 4.  –  power must be 0
term = 6C4 × 2^2 × (−1)^4 put r = 4
term = 15 × 4 × 1 = 60 6C4 = 15

Answer: 60.

D2. Evaluate (99)³ using the Binomial Theorem.

▶ Reveal full working

Write 99 as 100 − 1 and expand the cube; the minus sign makes alternate terms negative.

(99)^3 = (100 − 1)^3 convenient split
  = 100^3 − 3(100)^2(1) + 3(100)(1)^2 − 1 expand the cube
  = 1000000 − 30000 + 300 − 1 work out each term
  = 970299 add and subtract

Answer: 970299.

D3. In (1 + x)ⁿ, the coefficients of the 2nd and 3rd terms are in the ratio 1 : 2. Find n.

▶ Reveal full working

The 2nd and 3rd coefficients are nC1 and nC2. Form the ratio and solve.

nC1 : nC2 = 1 : 2 given ratio
nC1 / nC2 = 1/2 write as a fraction
n ÷ (n(n−1)/2) = 1/2 substitute the formulae
2/(n − 1) = 1/2 simplify the left side
n − 1 = 4 cross-multiply
n = 5 add 1

Answer: n = 5.

D4. Using the first three terms of the binomial expansion, estimate (1.01)⁵.

▶ Reveal full working

Write 1.01 as 1 + 0.01 and keep only the first three terms, since later terms are tiny.

(1.01)^5 = (1 + 0.01)^5 rewrite the base
  = 1 + 5(0.01) + 10(0.01)^2 + … first three binomial terms
  = 1 + 0.05 + 0.001 evaluate each term
  = 1.051 add (later terms are negligible)

Answer: About 1.051.

Chapter Summary

Everything in One Glance

The Theorem

(a + b)ⁿ expands into (n + 1) terms; the power of a falls from n to 0 while the power of b rises from 0 to n.

  

Pascal’s Triangle

Each row gives the coefficients; every inside number is the sum of the two above it, by Pascal’s rule.
 

Coefficients

The coefficient of the bᵣ term is ⁿCᵣ = n!/(r!(n − r)!), and ⁿCᵣ = ⁿCₙ₋ᵣ makes each row symmetric.

  

General Term

Tₛ₊₁ = ⁿCᵣ aⁿ₋ᵣ bᵣ. Set r one less than the term number to find any single term.
 

Middle Term

Even n: one middle term, the (n/2 + 1)th. Odd n: two middle terms, the ((n+1)/2)th and ((n+3)/2)th.

  

Independent of x

Collect the powers of x, set the total power to 0, solve for r, and substitute back to get the constant term.
 
8

Are You Exam-Ready?

8-Point Exam Quick-Check

1 How many terms are in the expansion of (a + b)¹⁰?
 
2 Write the row of Pascal’s triangle for n = 5.
 
3 State the general term of (a + b)ⁿ.
 
4 Find the 4th term in the expansion of (x + 1)⁶.
 
5 Find the middle term of (x + y)₈.
 
6 Find the term independent of x in (x + 1/x)⁴.
 
7 In (1 + x)⁷, what is the sum of all the coefficients?
 
8 Expand (a − b)³.
 

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Class 11 Mathematics Chapter 7: Binomial Theorem, Complete Notes and Practice

These free Class 11 Maths Chapter 7 Binomial Theorem notes follow the latest NCERT 2026-27 syllabus and give complete, exam-ready coverage for board exams and competitive foundations. Includes worked examples, step-by-step proofs and graded practice, free on SchoolRevise.com.
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