class 11 notes

Class 11 Mathematics Notes Chapter 8 (Binomial theorem) – Mathematics Book

Philoid

Philoid

— 6 min read

Mathematics
Alright class, let's get started with a very important chapter for your exams – Chapter 8: Binomial Theorem from your NCERT Class 11 Mathematics textbook. This theorem provides a powerful tool for expanding binomial expressions raised to any positive integer power, and understanding its nuances is crucial.

Chapter 8: Binomial Theorem - Detailed Notes

1. Introduction

  • A binomial is an algebraic expression containing two terms (e.g., a + b, x - 2y).
  • The Binomial Theorem gives a formula for the expansion of (a + b)ⁿ where 'n' is any positive integer (and can be extended to other exponents, though we focus on positive integers here as per NCERT Class 11).
  • We know simple expansions:
    • (a + b)⁰ = 1
    • (a + b)¹ = a + b
    • (a + b)² = a² + 2ab + b²
    • (a + b)³ = a³ + 3a²b + 3ab² + b³
  • The theorem provides a systematic way to find the coefficients and terms for higher powers without tedious multiplication.

2. Binomial Theorem for Positive Integral Index 'n'

The statement of the theorem is:
For any positive integer 'n',
(a + b)ⁿ = ⁿC₀ aⁿ b⁰ + ⁿC₁ aⁿ⁻¹ b¹ + ⁿC₂ aⁿ⁻² b² + ... + ⁿCᵣ aⁿ⁻ʳ bʳ + ... + ⁿCₙ a⁰ bⁿ

This can be written concisely using summation notation:
(a + b)ⁿ = Σ_{r=0}^{n} ⁿCᵣ aⁿ⁻ʳ bʳ

Where:

  • n is a positive integer (the exponent).
  • a and b are the terms in the binomial (can be real numbers, complex numbers, or even expressions).
  • r is the term index, starting from 0.
  • ⁿCᵣ (read as "n choose r") are the Binomial Coefficients.

3. Binomial Coefficients (ⁿCᵣ)

  • ⁿCᵣ represents the number of ways to choose 'r' items from a set of 'n' distinct items.
  • Formula: ⁿCᵣ = n! / [ r! * (n - r)! ]
    • where '!' denotes the factorial (e.g., 5! = 5 × 4 × 3 × 2 × 1 = 120).
    • Remember: 0! = 1.
  • Important Properties of ⁿCᵣ:
    • ⁿC₀ = 1 (Choosing 0 items)
    • ⁿCₙ = 1 (Choosing all n items)
    • ⁿC₁ = n
    • ⁿC_(n-1) = n
    • Symmetry: ⁿCᵣ = ⁿC_(n-r) (e.g., ¹⁰C₂ = ¹⁰C₈)
    • Pascal's Rule: ⁿCᵣ + ⁿC_(r-1) = ⁿ⁺¹Cᵣ (This is the basis for Pascal's Triangle)

4. Pascal's Triangle

  • This is a triangular array of numbers where each number is the sum of the two numbers directly above it.
  • It provides the binomial coefficients for expansions.
        1             (n=0) --> ⁰C₀
       1 1            (n=1) --> ¹C₀ ¹C₁
      1 2 1           (n=2) --> ²C₀ ²C₁ ²C₂
     1 3 3 1          (n=3) --> ³C₀ ³C₁ ³C₂ ³C₃
    1 4 6 4 1         (n=4) --> ⁴C₀ ⁴C₁ ⁴C₂ ⁴C₃ ⁴C₄
   1 5 10 10 5 1      (n=5) --> ⁵C₀ ⁵C₁ ⁵C₂ ⁵C₃ ⁵C₄ ⁵C₅
  ... and so on
  • The numbers in the (n+1)th row correspond to the coefficients ⁿC₀, ⁿC₁, ..., ⁿCₙ in the expansion of (a + b)ⁿ.

5. Important Observations/Properties of the Binomial Expansion (a + b)ⁿ

  • Number of Terms: The expansion has (n + 1) terms.
  • Sum of Indices: In each term, the sum of the powers of 'a' and 'b' is always equal to 'n' (i.e., (n - r) + r = n).
  • Coefficients: The coefficients (ⁿCᵣ) are equidistant from the beginning and the end are equal (due to ⁿCᵣ = ⁿC_(n-r)).
  • Sum of Coefficients: If we put a = 1 and b = 1 in the expansion, we get:
    (1 + 1)ⁿ = ⁿC₀ + ⁿC₁ + ⁿC₂ + ... + ⁿCₙ
    2ⁿ = Σ_{r=0}^{n} ⁿCᵣ
  • Alternating Sum of Coefficients: If we put a = 1 and b = -1, we get:
    (1 - 1)ⁿ = ⁿC₀ - ⁿC₁ + ⁿC₂ - ⁿC₃ + ... + (-1)ⁿ ⁿCₙ
    0 = Σ_{r=0}^{n} (-1)ʳ ⁿCᵣ (for n ≥ 1)
    This implies: ⁿC₀ + ⁿC₂ + ⁿC₄ + ... = ⁿC₁ + ⁿC₃ + ⁿC₅ + ... = 2ⁿ⁻¹ (Sum of even coefficients = Sum of odd coefficients)

6. General Term (T_(r+1))

  • The (r + 1)th term in the expansion of (a + b)ⁿ is called the General Term. It's extremely useful for finding specific terms.
  • Formula: T_(r+1) = ⁿCᵣ aⁿ⁻ʳ bʳ
    • Crucial: Remember this is the (r+1)th term, involving ⁿCᵣ. So, the 5th term corresponds to r = 4, the 10th term to r = 9, etc.

7. Middle Term(s)

The middle term depends on whether 'n' is even or odd. The total number of terms is (n+1).

  • Case 1: n is Even

    • The number of terms (n + 1) is odd.
    • There is only one middle term.
    • It is the (n/2 + 1)th term.
    • Middle Term = T_((n/2) + 1) = ⁿC_(n/2) * a^(n/2) * b^(n/2)

  • Case 2: n is Odd

    • The number of terms (n + 1) is even.
    • There are two middle terms.
    • They are the ((n+1)/2)th term and the ((n+1)/2 + 1)th term.
    • Middle Term 1 = T_((n+1)/2) = ⁿC_((n-1)/2) * a^((n+1)/2) * b^((n-1)/2)
    • Middle Term 2 = T_(((n+1)/2) + 1) = ⁿC_((n+1)/2) * a^((n-1)/2) * b^((n+1)/2)
    • Note: For odd n, the coefficients of the two middle terms are equal: ⁿC_((n-1)/2) = ⁿC_((n+1)/2).

8. Finding a Specific Term / Term Independent of x

  • To find the k-th term: Use the general term formula T_(r+1) with r = k - 1.
  • To find the term independent of x (or constant term):
    1. Write down the general term T_(r+1) = ⁿCᵣ aⁿ⁻ʳ bʳ for the given expansion (e.g., (px^p + q/x^q)ⁿ).
    2. Simplify the expression to collect all powers of x, say as x^k, where k will be an expression involving 'r'.
    3. Set the exponent of x to zero (k = 0).
    4. Solve the equation k = 0 for 'r'. 'r' must be a non-negative integer between 0 and n (inclusive).
    5. If a valid integer 'r' is found, substitute this value back into the T_(r+1) formula to get the term independent of x. If no such integer 'r' exists, there is no term independent of x.

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

  1. a = x², b = 1/x = x⁻¹, n = 9.
  2. T_(r+1) = ⁹Cᵣ (x²)⁹⁻ʳ (x⁻¹)ʳ = ⁹Cᵣ x^(18-2r) x⁻ʳ = ⁹Cᵣ x^(18-3r)
  3. Set exponent of x to 0: 18 - 3r = 0
  4. Solve for r: 3r = 18 => r = 6. (This is valid as 0 ≤ 6 ≤ 9).
  5. Substitute r = 6 back: T_(6+1) = T₇ = ⁹C₆ x^(18-3*6) = ⁹C₆ x⁰ = ⁹C₆.
    ⁹C₆ = ⁹C₃ = (9 × 8 × 7) / (3 × 2 × 1) = 3 × 4 × 7 = 84.
    So, the term independent of x is 84.

Key Takeaways for Exams:

  • Memorize the Binomial Theorem statement and the General Term formula (T_(r+1)).
  • Be proficient in calculating ⁿCᵣ and know its properties (especially symmetry).
  • Understand how to find the number of terms and the middle term(s).
  • Practice finding specific terms (e.g., 4th term, term containing x⁵).
  • Master the technique for finding the term independent of x.
  • Remember the sum of coefficients (2ⁿ) and the alternating sum (0).

Multiple Choice Questions (MCQs)

Here are 10 MCQs based on the concepts discussed. Try to solve them yourself first!

  1. The total number of terms in the expansion of (2x + 3y)¹⁰ is:
    a) 10
    b) 11
    c) 12
    d) 20

  2. The value of ⁷C₃ is:
    a) 21
    b) 35
    c) 42
    d) 7

  3. The coefficient of the middle term in the expansion of (1 + x)⁸ is:
    a) ⁸C₃
    b) ⁸C₄
    c) ⁸C₅
    d) ⁸C₆

  4. The 4th term in the expansion of (x - 2y)⁷ is:
    a) -280 x⁴ y³
    b) 280 x⁴ y³
    c) -560 x³ y⁴
    d) 560 x³ y⁴

  5. The sum of the coefficients in the expansion of (x - 2y + 3z)⁵ is: (Hint: Think about substituting values for variables)
    a) 32
    b) 1
    c) 243
    d) 32

  6. If ⁿC₈ = ⁿC₆, then the value of n is:
    a) 8
    b) 6
    c) 14
    d) 2

  7. The term independent of x in the expansion of (x + 1/x)⁶ is:
    a) 15
    b) 20
    c) 6
    d) 1

  8. The coefficients of the two middle terms in the expansion of (a + x)⁹ are:
    a) ⁹C₃ and ⁹C₆
    b) ⁹C₄ and ⁹C₅
    c) ⁹C₅ and ⁹C₆
    d) ⁹C₄ and ⁹C₄

  9. The value of ⁵C₀ + ⁵C₁ + ⁵C₂ + ⁵C₃ + ⁵C₄ + ⁵C₅ is:
    a) 16
    b) 31
    c) 32
    d) 64

  10. The general term (T_(r+1)) in the expansion of (3x² - 1/x)¹² is:
    a) ¹²Cᵣ (3x²)¹²⁻ʳ (-1/x)ʳ
    b) ¹²Cᵣ (3x²)ʳ (-1/x)¹²⁻ʳ
    c) ¹²C_(r+1) (3x²)¹²⁻ʳ (-1/x)ʳ
    d) ¹²Cᵣ (3x²)¹²⁻ʳ (1/x)ʳ

Answers to MCQs:

  1. b) 11 (Number of terms = n + 1 = 10 + 1 = 11)
  2. b) 35 (⁷C₃ = 7! / (3! * 4!) = (7 × 6 × 5) / (3 × 2 × 1) = 35)
  3. b) ⁸C₄ (n=8 is even. Middle term is (8/2 + 1) = 5th term. T₅ = T_(4+1) = ⁸C₄ (1)⁴ (x)⁴. Coefficient is ⁸C₄)
  4. a) -280 x⁴ y³ (T₄ = T_(3+1) = ⁷C₃ x⁷⁻³ (-2y)³ = 35 * x⁴ * (-8y³) = -280 x⁴ y³)
  5. d) 32 (Put x=1, y=1, z=1. (1 - 2(1) + 3(1))⁵ = (1 - 2 + 3)⁵ = 2⁵ = 32)
  6. c) 14 (Using ⁿCᵣ = ⁿC_(n-r), we have ⁿC₈ = ⁿC_(n-8). So, n-8 = 6 or 8 = 6 (not possible). Or using ⁿCₓ = ⁿC<0xE1><0xB5><0xA7>, then x + y = n. So, 8 + 6 = n => n = 14)
  7. b) 20 (T_(r+1) = ⁶Cᵣ x⁶⁻ʳ (1/x)ʳ = ⁶Cᵣ x⁶⁻²ʳ. Set 6 - 2r = 0 => r = 3. Term is ⁶C₃ = (6 × 5 × 4) / (3 × 2 × 1) = 20)
  8. b) ⁹C₄ and ⁹C₅ (n=9 is odd. Middle terms are ((9+1)/2) = 5th and ((9+1)/2 + 1) = 6th. T₅ = T_(4+1) has coefficient ⁹C₄. T₆ = T_(5+1) has coefficient ⁹C₅. Note: ⁹C₄ = ⁹C₅)
  9. c) 32 (Sum of coefficients = 2ⁿ = 2⁵ = 32)
  10. a) ¹²Cᵣ (3x²)¹²⁻ʳ (-1/x)ʳ (Direct application of T_(r+1) = ⁿCᵣ aⁿ⁻ʳ bʳ with a=3x², b=-1/x, n=12)

Revise these notes thoroughly, practice problems from your NCERT book and previous year question papers. Let me know if any specific part needs more clarification! Good luck with your preparation.

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