Binomial Theorem – Class 11 Maths Chapter 7 Notes & NCERT PDF

Class 11MathematicsChapter 7NCERT book: Mathematics - Class 11

Notes and a simple summary of Chapter 7 of the NCERT Class 11 Mathematics book, with the official chapter PDF, flashcards, an MCQ quiz and an AI tutor for your doubts.

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Chapter 7: Binomial Theorem

Introduction

Squaring or cubing 98 or 999 is easy using (100 − 2)² or (1000 − 1)³, but (98)⁵ or (101)⁶ needs repeated multiplication. The binomial theorem gives a direct expansion of (a + b)ⁿ for any positive integer n, so such powers can be found quickly. The chapter (rationalised syllabus) covers the theorem for positive integral indices, Pascal's triangle and simple applications.

Key Concepts & Theorems

7.2 Pattern in the expansions

(a + b)⁰ = 1; (a + b)¹ = a + b; (a + b)² = a² + 2ab + b²; (a + b)³ = a³ + 3a²b + 3ab² + b³; (a + b)⁴ = a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴.

The book's three observations:

(i) The number of terms is one more than the index.

(ii) The power of the first quantity a decreases by 1 in each successive term, while the power of b increases by 1.

(iii) In every term, the sum of the indices of a and b equals the index n.

Pascal's triangle

Writing the coefficients row by row gives 1; 1 1; 1 2 1; 1 3 3 1; 1 4 6 4 1; … Each entry is the sum of the two numbers just above it, and every row begins and ends with 1. This array is Pascal's triangle, also known as Meru Prastara (after Pingala). Since the entries are ⁿC₀, ⁿC₁, …, ⁿCₙ, the row for index 7 can be written directly: 1, 7, 21, 35, 35, 21, 7, 1.

Using the row for index 5: (2x + 3y)⁵ = 32x⁵ + 240x⁴y + 720x³y² + 1080x²y³ + 810xy⁴ + 243y⁵.

Binomial Theorem for any positive integer n

(a + b)ⁿ = ⁿC₀aⁿ + ⁿC₁aⁿ⁻¹b + ⁿC₂aⁿ⁻²b² + … + ⁿCₙ₋₁abⁿ⁻¹ + ⁿCₙbⁿ

In sigma notation, (a + b)ⁿ = Σ ⁿCₖ aⁿ⁻ᵏ bᵏ for k = 0 to n. The book proves it by the principle of mathematical induction, using ᵏCᵣ + ᵏCᵣ₋₁ = ᵏ⁺¹Cᵣ. The coefficients ⁿCᵣ are the binomial coefficients.

Illustration: (x + 2)⁶ = x⁶ + 12x⁵ + 60x⁴ + 160x³ + 240x² + 192x + 64.

Special cases

(x − y)ⁿ = ⁿC₀xⁿ − ⁿC₁xⁿ⁻¹y + ⁿC₂xⁿ⁻²y² − … + (−1)ⁿ ⁿCₙyⁿ — signs alternate. Example: (x − 2y)⁵ = x⁵ − 10x⁴y + 40x³y² − 80x²y³ + 80xy⁴ − 32y⁵.

(1 + x)ⁿ = ⁿC₀ + ⁿC₁x + ⁿC₂x² + … + ⁿCₙxⁿ. Putting x = 1: 2ⁿ = ⁿC₀ + ⁿC₁ + … + ⁿCₙ.

(1 − x)ⁿ = ⁿC₀ − ⁿC₁x + ⁿC₂x² − … + (−1)ⁿ ⁿCₙxⁿ. Putting x = 1: 0 = ⁿC₀ − ⁿC₁ + ⁿC₂ − … + (−1)ⁿ ⁿCₙ.

Worked Examples (from the chapter)

Example 1: (x² + 3/x)⁴ = ⁴C₀(x²)⁴ + ⁴C₁(x²)³(3/x) + ⁴C₂(x²)²(3/x)² + ⁴C₃(x²)(3/x)³ + ⁴C₄(3/x)⁴ = x⁸ + 12x⁵ + 54x² + 108/x + 81/x⁴.

Example 2: (98)⁵ = (100 − 2)⁵ = 10000000000 − 1000000000 + 40000000 − 800000 + 8000 − 32 = 9039207968.

Example 3: Which is larger, (1.01)¹⁰⁰⁰⁰⁰⁰ or 10,000? (1 + 0.01)¹⁰⁰⁰⁰⁰⁰ = 1 + 1000000 × 0.01 + (other positive terms) = 1 + 10000 + … > 10000. So (1.01)¹⁰⁰⁰⁰⁰⁰ is larger.

Example 4: Prove 6ⁿ − 5n always leaves remainder 1 when divided by 25. (1 + 5)ⁿ = 1 + 5n + 5²·ⁿC₂ + 5³·ⁿC₃ + … + 5ⁿ, so 6ⁿ − 5n = 1 + 25(ⁿC₂ + 5·ⁿC₃ + … + 5ⁿ⁻²) = 25k + 1.

Extra practice (Exercise 7.1 type): (102)⁵ = (100 + 2)⁵ = 11040808032; (99)⁵ = (100 − 1)⁵ = 9509900499; (101)⁴ = 104060401; (96)³ = (100 − 4)³ = 884736. Also 9ⁿ⁺¹ − 8n − 9 is divisible by 64: write 9ⁿ⁺¹ = (1 + 8)ⁿ⁺¹ and expand.

How the induction proof works (in words)

Base step: for n = 1, (a + b)¹ = ¹C₀a + ¹C₁b = a + b, so the statement P(1) is true.

Inductive step: assume P(k) is true, i.e. (a + b)ᵏ = ᵏC₀aᵏ + ᵏC₁aᵏ⁻¹b + … + ᵏCₖbᵏ.

Multiply both sides by (a + b). Multiplying by a raises every power of a by one; multiplying by b raises every power of b by one.

Group the like terms: each middle coefficient becomes a sum such as ᵏC₁ + ᵏC₀.

Use ᵏCᵣ + ᵏCᵣ₋₁ = ᵏ⁺¹Cᵣ (Theorem 6 of the previous chapter) together with ᵏC₀ = 1 = ᵏ⁺¹C₀ and ᵏCₖ = 1 = ᵏ⁺¹Cₖ₊₁. The result is exactly the formula for n = k + 1.

So P(k) true ⇒ P(k + 1) true, and by the principle of mathematical induction the theorem holds for every positive integer n.

Two more expansions (Exercise 7.1 style, fully simplified)

(1 − 2x)⁵ = 1 − 10x + 40x² − 80x³ + 80x⁴ − 32x⁵. Here a = 1 and b = −2x, so the signs alternate and each coefficient is ⁵Cᵣ × 2ʳ.

(2x − 3)⁶ = 64x⁶ − 576x⁵ + 2160x⁴ − 4320x³ + 4860x² − 2916x + 729. Check: the last term is (−3)⁶ = 729 and the coefficient of x⁵ is ⁶C₁ × 2⁵ × (−3) = −576.

(x + 1)⁶ + (x − 1)⁶ = 2x⁶ + 30x⁴ + 30x² + 2 — the odd-power terms cancel, which is the idea behind Exercise 7.1, Q12.

Formula Sheet

(a + b)ⁿ = Σ ⁿCₖ aⁿ⁻ᵏ bᵏ, k = 0 to n

Number of terms = n + 1

(k + 1)th term (general term) has coefficient ⁿCₖ with aⁿ⁻ᵏbᵏ

ⁿCᵣ = ⁿCₙ₋ᵣ (coefficients equidistant from the ends are equal)

Sum of coefficients: ⁿC₀ + ⁿC₁ + … + ⁿCₙ = 2ⁿ

Alternating sum: ⁿC₀ − ⁿC₁ + ⁿC₂ − … = 0

Pascal: ⁿCᵣ + ⁿCᵣ₋₁ = ⁿ⁺¹Cᵣ

Exam Focus

Expand (Exercise 7.1, Q1–5): write every ⁿCᵣ, then simplify powers carefully, keeping track of signs and powers of x in the denominator.

Evaluate (Q6–9): split into 100 ± small number.

Compare (Q10): keep the first two terms and argue that the rest are positive.

Divisibility (Q13): write the base as 1 + (divisor-related number) and expand.

Answer outline — "Find (a + b)⁴ − (a − b)⁴": terms with odd powers of b double, even ones cancel ⇒ 8a³b + 8ab³ = 8ab(a² + b²).

Answer outline — "Which is larger, (1.1)¹⁰⁰⁰⁰ or 1000?": (1 + 0.1)¹⁰⁰⁰⁰ > 1 + 10000 × 0.1 = 1001 > 1000.

Key Terms

Binomial (द्विपद) — expression with two terms

Binomial theorem (द्विपद प्रमेय)

Index / exponent (घातांक)

Binomial coefficient (द्विपद गुणांक) — ⁿCᵣ

Pascal's triangle / Meru Prastara (पास्कल त्रिभुज / मेरु प्रस्तार)

Expansion (प्रसार)

Term (पद)

Mathematical induction (गणितीय आगमन)

Sigma notation (सिग्मा संकेतन)

Remainder (शेषफल)

Common Mistakes

Forgetting that there are n + 1 terms, not n.

Losing the alternating signs in (a − b)ⁿ.

Not raising the coefficient to the power: (2x)⁵ = 32x⁵, not 2x⁵.

Mishandling negative powers: (3/x)³ = 27/x³.

Stopping the approximation too early when both "other terms" could be negative.

Mixing up Pascal's row index: the row 1 4 6 4 1 is for n = 4.

💡 Key Learning Points

✓Describe the pattern of (a + b)ⁿ: n + 1 terms, falling powers of a, rising powers of b, and index sum n in every term.
✓Build and use Pascal's triangle (Meru Prastara) and identify its entries as ⁿCᵣ.
✓State the binomial theorem (a + b)ⁿ = Σ ⁿCₖaⁿ⁻ᵏbᵏ for positive integers n and understand its proof by induction.
✓Expand (x − y)ⁿ, (1 + x)ⁿ and (1 − x)ⁿ, and derive Σ ⁿCᵣ = 2ⁿ and the alternating sum 0.
✓Apply the theorem to evaluate powers like (98)⁵, compare numbers and prove divisibility results.

👨‍🏫 Teaching Tips

→Ask students to compute (98)² mentally with (100 − 2)², then pose (98)⁵ to create the need for the theorem.
→Let students build Pascal's triangle to row 7 themselves and discover the 'add the two above' rule.
→Link Pascal's triangle to the Indian name Meru Prastara and to ⁿCᵣ from Chapter 6.
→Colour-code the powers of a and b in an expansion to make observations (ii) and (iii) visible.
→Practise with alternating signs using (x − 2y)⁵ before tougher fractional expansions.
→Show the 6ⁿ − 5n remainder proof numerically for n = 1, 2, 3 before the general argument.

📋 Assessment Questions

Can the student write the full expansion of an expression like (2x − 3)⁶ correctly?
Can the student write any row of Pascal's triangle using ⁿCᵣ?
Can the student evaluate a number such as (102)⁵ using the theorem?
Can the student prove a divisibility statement like 9ⁿ⁺¹ − 8n − 9 divisible by 64?
Can the student use special cases to find Σ ⁿCᵣ?

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