CBSE Class 11 Mathematics Chapter 7 Notes: Binomial Theorem

Chapter Overview

Binomial Theorem is Chapter 7 of CBSE Class 11 Mathematics and belongs to the Algebra unit (Unit II). It provides a formula to expand expressions of the form (a + b)ⁿ without manual multiplication. This theorem is extensively used in calculus, probability (binomial distribution), and approximations in physics and engineering.

This guide covers every NCERT topic with the theorem, formulae, Pascal's triangle, properties, worked examples, common mistakes, and exam tips.

The Binomial Theorem

Statement (Positive Integral Index)

For any positive integer n:

(a + b)ⁿ = ⁿC₀ aⁿ + ⁿC₁ aⁿ⁻¹ b + ⁿC₂ aⁿ⁻² b² + … + ⁿCᵣ aⁿ⁻ʳ bʳ + … + ⁿCₙ bⁿ

where ⁿCr = n!/[r!(n − r)!] are the binomial coefficients.

Key Features of the Expansion

General and Middle Terms

General Term

The (r + 1)th term in the expansion of (a + b)ⁿ is:

T_{r+1} = ⁿCr aⁿ⁻ʳ bʳ

where r = 0, 1, 2, …, n.

• First term: T₁ = ⁿC₀ aⁿ = aⁿ (r = 0)
• Last term: T_{n+1} = ⁿCₙ bⁿ = bⁿ (r = n)

Middle Terms

Examples:

• (a + b)⁶: n = 6 (even) → middle term is T₄ (r = 3): ⁶C₃ a³b³
• (a + b)⁷: n = 7 (odd) → middle terms are T₄ (r = 3) and T₅ (r = 4)

Pascal's Triangle

Pascal's triangle arranges binomial coefficients in a triangular pattern:

Each entry is the sum of the two entries directly above it: ⁿCr = ⁽ⁿ⁻¹⁾C(r−1) + ⁽ⁿ⁻¹⁾Cr.

Pascal's Triangle Rows

Properties of Binomial Coefficients

Deriving Key Properties

Sum of all coefficients: Set a = 1, b = 1 in (a + b)ⁿ: (1 + 1)ⁿ = 2ⁿ = ⁿC₀ + ⁿC₁ + … + ⁿCₙ

Sum of even = Sum of odd: Set a = 1, b = −1: (1 − 1)ⁿ = 0 = ⁿC₀ − ⁿC₁ + ⁿC₂ − ⁿC₃ + … This means (ⁿC₀ + ⁿC₂ + …) − (ⁿC₁ + ⁿC₃ + …) = 0 Since the total is 2ⁿ and both halves are equal, each sum = 2ⁿ⁻¹.

Greatest Binomial Coefficient

The binomial coefficients increase up to the middle term(s) and then decrease symmetrically.

Finding the Greatest Term

To find the greatest term in (a + b)ⁿ, compute:

If r is an integer, then T_r and T_{r+1} are equal and both are the greatest terms. Otherwise, T_{r+1} is the greatest term.

Important Numericals

1. Expanding Using the Binomial Theorem

Expand (x + 2)⁵.

Using (a + b)⁵ = ⁵C₀a⁵ + ⁵C₁a⁴b + ⁵C₂a³b² + ⁵C₃a²b³ + ⁵C₄ab⁴ + ⁵C₅b⁵:

(x + 2)⁵ = x⁵ + 5x⁴(2) + 10x³(4) + 10x²(8) + 5x(16) + 32 = x⁵ + 10x⁴ + 40x³ + 80x² + 80x + 32

2. Finding a Specific Term

Find the 4th term in the expansion of (2x − 3)⁷.

T₄ = T_{r+1} where r = 3. T₄ = ⁷C₃ (2x)⁷⁻³ (−3)³ = 35 × (2x)⁴ × (−27) = 35 × 16x⁴ × (−27) = −15120x⁴

3. Finding the Middle Term

Find the middle term in the expansion of (3x − 2y)⁸.

n = 8 (even). Middle term = T_{(8/2)+1} = T₅ (r = 4). T₅ = ⁸C₄ (3x)⁴ (−2y)⁴ = 70 × 81x⁴ × 16y⁴ = 90720x⁴y⁴

4. Term Independent of x

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

General term: T_{r+1} = ⁶Cr x^{6−r} (1/x)^r = ⁶Cr x^{6−2r}

For the term independent of x: 6 − 2r = 0 → r = 3.

T₄ = ⁶C₃ = 20

5. Coefficient of a Specific Power

Find the coefficient of x⁵ in the expansion of (1 + x)¹⁰.

T_{r+1} = ¹⁰Cr xʳ. We need r = 5. Coefficient = ¹⁰C₅ = 252.

6. Finding the Greatest Coefficient

Find the greatest coefficient in the expansion of (a + b)⁹.

n = 9 (odd). Greatest coefficients: ⁹C₄ and ⁹C₅. ⁹C₄ = 9!/(4! × 5!) = (9 × 8 × 7 × 6)/(4 × 3 × 2 × 1) = 126

7. Numerical Approximation

Find (1.02)⁶ using the binomial theorem.

(1.02)⁶ = (1 + 0.02)⁶ = ⁶C₀(1)⁶ + ⁶C₁(1)⁵(0.02) + ⁶C₂(1)⁴(0.02)² + ⁶C₃(1)³(0.02)³ + … = 1 + 6(0.02) + 15(0.0004) + 20(0.000008) + … = 1 + 0.12 + 0.006 + 0.00016 + … ≈ 1.12616

Binomial Theorem for Any Index (Overview)

For fractional or negative n, the binomial expansion becomes an infinite series. The formula uses generalised binomial coefficients but this is beyond the CBSE Class 11 computation scope. Key points to know:

• The expansion (1 + x)ⁿ = 1 + nx + n(n−1)x²/2! + n(n−1)(n−2)x³/3! + … is valid when |x| < 1.
• The series has infinitely many terms (unlike positive integer n where it terminates).
• This is used for approximations: (1 + x)ⁿ ≈ 1 + nx when x is very small.

Practice Problems

Problem 1

Expand (2x − 1)⁵ using the binomial theorem.

(2x − 1)⁵ = ⁵C₀(2x)⁵ + ⁵C₁(2x)⁴(−1) + ⁵C₂(2x)³(−1)² + ⁵C₃(2x)²(−1)³ + ⁵C₄(2x)(−1)⁴ + ⁵C₅(−1)⁵ = 32x⁵ − 80x⁴ + 80x³ − 40x² + 10x − 1

Problem 2

Find the coefficient of x⁴ in the expansion of (1 + x + x²)¹⁰.

Write (1 + x + x²)¹⁰ = (1 + x(1 + x))¹⁰. The general term is ¹⁰Cr xʳ(1 + x)ʳ. Within (1 + x)ʳ, the general term is ʳCk xᵏ. So the combined term is ¹⁰Cr × ʳCk × x^(r+k). We need r + k = 4.

Total coefficient = 45 + 360 + 210 = 615

• r = 0, k = 4: ¹⁰C₀ × ⁰C₄ = 0 (impossible since ⁰C₄ = 0)
• r = 1, k = 3: ¹⁰C₁ × ¹C₃ = 0
• r = 2, k = 2: ¹⁰C₂ × ²C₂ = 45 × 1 = 45
• r = 3, k = 1: ¹⁰C₃ × ³C₁ = 120 × 3 = 360
• r = 4, k = 0: ¹⁰C₄ × ⁴C₀ = 210 × 1 = 210

Problem 3

Which term in the expansion of (x − 2y)⁸ has the greatest absolute coefficient?

The binomial coefficients ⁸Cr are maximised at r = 4: ⁸C₄ = 70. The corresponding term is T₅ = ⁸C₄ x⁴(−2y)⁴ = 70 × x⁴ × 16y⁴ = 1120x⁴y⁴.

Problem 4

If the coefficients of x² and x³ in (1 + x)ⁿ are in the ratio 5 : 8, find n.

ⁿC₂/ⁿC₃ = 5/8 [n!/(2!(n−2)!)] / [n!/(3!(n−3)!)] = 5/8 [3!(n−3)!] / [2!(n−2)!] = 5/8 3/(n−2) = 5/8 24 = 5(n − 2) 24 = 5n − 10 n = 34/5 — not an integer, so recheck. Actually: 3 × 8 = 5(n − 2) → 24 = 5n − 10 → 34 = 5n → n = 34/5. Since n must be a positive integer, there may be no solution with this exact ratio. However, the NCERT problem uses ratio 1 : 7 giving n = 8.

NCERT Exercise Mapping

Common Mistakes

1. Confusing term number with r value: The 4th term corresponds to r = 3 (since T_{r+1}). The most common error is setting r = 4 when asked for the 4th term. 2. Forgetting the sign when b is negative: In (a − b)ⁿ, odd powers of b carry a negative sign. For example, in (x − y)⁵, T₃ = ¹⁰C₂x³y² is positive, but T₄ = ¹⁰C₃x²y³ is negative. 3. Wrong middle term for odd n: When n is odd, there are TWO middle terms. Writing only one loses marks. 4. Setting the wrong equation for "term independent of x": You must set the NET exponent of x to zero in the general term. Students sometimes set only one part's exponent to zero. 5. Not converting ⁿCr to a numerical value: In exams, always compute ⁿCr as a number. Writing ⁶C₃ as the final answer instead of 20 loses the evaluation mark. 6. Forgetting that the expansion has n + 1 terms, not n terms: (a + b)⁵ has 6 terms (0 through 5), not 5 terms. 7. Misapplying the binomial theorem to fractional or negative n: The standard formula with ⁿCr as factorial applies only for positive integers. For fractional/negative n, the expansion is an infinite series with different coefficient rules (not in the Class 11 NCERT scope for computation).

Exam Tips

• 1-mark questions: State the binomial theorem, write the general term, or give a row of Pascal's triangle.
• 2-mark questions: Expand (a + b)ⁿ for small n (3 or 4), or find a specific term using the general term formula.
• 4-mark questions: Find the middle term(s), find the term independent of x, or compute the coefficient of a specific power of x.
• 6-mark questions: Prove a property of binomial coefficients, find the greatest term using the ratio method, or use the binomial theorem for numerical approximation.
• Always write the general term formula before substituting: T_{r+1} = ⁿCr aⁿ⁻ʳ bʳ.
• When b is negative, write (−b)ʳ explicitly and track the sign.
• For "coefficient of x⁵" questions, find r first, then compute ⁿCr.
• For numerical approximations, use only the first few terms — higher-order terms are negligible for small values.
• Verify your answer by checking that the sum of coefficients equals 2ⁿ (set a = b = 1).

Quick Revision Notes

• (a + b)ⁿ = Σ ⁿCr aⁿ⁻ʳ bʳ for r = 0 to n. Total terms = n + 1.
• General term: T_{r+1} = ⁿCr aⁿ⁻ʳ bʳ. The rth term uses r − 1 in the formula.
• Middle term(s): n even → one middle term T_{(n/2)+1}; n odd → two middle terms.
• Pascal's triangle: each entry = sum of two above. Row n sums to 2ⁿ.
• ⁿC₀ + ⁿC₁ + … + ⁿCₙ = 2ⁿ. Even-indexed sum = odd-indexed sum = 2ⁿ⁻¹.
• ⁿCr = ⁿC(n−r). Symmetry of coefficients.
• ⁿCr + ⁿC(r−1) = (n+1)Cr. Pascal's identity.
• Greatest coefficient: ⁿC(n/2) for even n; ⁿC((n−1)/2) = ⁿC((n+1)/2) for odd n.
• Term independent of x: set exponent of x to zero, solve for integer r.
• Negative b: track signs through (−b)ʳ — positive for even r, negative for odd r.
• Numerical approximation: (1 + x)ⁿ ≈ 1 + nx + n(n−1)x²/2 + … for small |x|.