CBSE Class 11 Mathematics Chapter 8: Binomial Theorem NCERT Solutions

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NCERT Solutions PDF Class 11 PDF

This chapter provides comprehensive NCERT Solutions for Class 11 Mathematics, focusing on the Binomial Theorem. Students will learn the fundamental principles of binomial expansion and how to apply them to various expressions. The solutions cover the expansion of binomials raised to positive integer powers using the binomial theorem formula. Key concepts include understanding binomial coefficients, the general term in an expansion, and applying the theorem to simplify complex expressions. These step-by-step solutions are designed to clarify the process of binomial expansion, helping students build a strong foundation in this important algebraic concept. They are ideal for exam preparation, offering clear explanations and accurate results for each exercise problem, ensuring students can confidently tackle binomial expansion questions in their assessments.

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BoardCBSE
ClassClass 11
SubjectMathematics
Session2026
LanguageEnglish
TypeNCERT Solutions
ChapterChapter 8: Binomial Theorem

Chapter summary

This chapter's NCERT Solutions for Class 11 Mathematics delve into the Binomial Theorem. It provides a systematic approach to expanding expressions of the form (a+b)^n, where n is a positive integer. The solutions guide students through calculating binomial coefficients and applying the expansion formula. Exercises focus on practical application, including expanding various binomial expressions with fractional and variable terms. This resource is crucial for mastering the mechanics of binomial expansion and preparing for related problems in higher mathematics.

Learning outcomes

  • Understand the Binomial Theorem for expanding expressions.
  • Apply the binomial theorem to expand expressions of the form (a+b)^n.
  • Calculate binomial coefficients using the combination formula.
  • Expand binomial expressions involving fractions and variables.
  • Simplify expanded binomial expressions.

Topics covered

Paper topics

  • Binomial Theorem
  • Expansion of binomials
  • Binomial coefficients
  • General term in binomial expansion
  • Combinations (nCr)
  • Algebraic expressions

Important topics

  • Binomial Theorem formula
  • Calculating binomial coefficients
  • Expanding (a+b)^n
  • Simplifying expanded terms

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Questions and Solutions

Question 1

Expand the expression (1-2x)^5.
Solution:

We use the Binomial Theorem to expand the given expression (1-2x)^5. The binomial theorem states that for any non-negative integer \(n\), the expansion of (a+b)^n is given by:

(a+b)^n = \sum_{r=0}^{n} {}^{n}C_{r} a^{n-r} b^{r}

In this case, \(a = 1\), \(b = -2x\), and \(n = 5\). Applying the formula:

(1-2x)^{5} = {}^{5}C_{0}(1)^{5}(-2x)^{0} + {}^{5}C_{1}(1)^{4}(-2x)^{1} + {}^{5}C_{2}(1)^{3}(-2x)^{2} + {}^{5}C_{3}(1)^{2}(-2x)^{3} + {}^{5}C_{4}(1)^{1}(-2x)^{4} + {}^{5}C_{5}(1)^{0}(-2x)^{5}

Now, we calculate the binomial coefficients:

{}^{5}C_{0} = 1, {}^{5}C_{1} = 5, {}^{5}C_{2} = \frac{5 \times 4}{2 \times 1} = 10, {}^{5}C_{3} = \frac{5 \times 4 \times 3}{3 \times 2 \times 1} = 10, {}^{5}C_{4} = 5, {}^{5}C_{5} = 1

Substitute these values back into the expansion:

= 1(1)(1) + 5(1)(-2x) + 10(1)(4x^2) + 10(1)(-8x^3) + 5(1)(16x^4) + 1(1)(-32x^5)

Simplify each term:

= 1 - 10x + 40x^2 - 80x^3 + 80x^4 - 32x^5

Thus, the expansion of (1-2x)^5 is 1 - 10x + 40x^2 - 80x^3 + 80x^4 - 32x^5.

Question 2

Expand the expression \left(\frac{2}{x} - \frac{x}{2}\right)^5.
Solution:

We apply the Binomial Theorem to expand \left(\frac{2}{x} - \frac{x}{2}\right)^5. Here, \(a = \frac{2}{x}\), \(b = -\frac{x}{2}\), and \(n = 5\). The expansion is:

\left(\frac{2}{x} - \frac{x}{2}\right)^{5} = {}^{5}C_{0}\left(\frac{2}{x}\right)^{5} + {}^{5}C_{1}\left(\frac{2}{x}\right)^{4}\left(-\frac{x}{2}\right)^{1} + {}^{5}C_{2}\left(\frac{2}{x}\right)^{3}\left(-\frac{x}{2}\right)^{2} + {}^{5}C_{3}\left(\frac{2}{x}\right)^{2}\left(-\frac{x}{2}\right)^{3} + {}^{5}C_{4}\left(\frac{2}{x}\right)^{1}\left(-\frac{x}{2}\right)^{4} + {}^{5}C_{5}\left(\frac{2}{x}\right)^{0}\left(-\frac{x}{2}\right)^{5}

Calculate the binomial coefficients: {}^{5}C_{0}=1, {}^{5}C_{1}=5, {}^{5}C_{2}=10, {}^{5}C_{3}=10, {}^{5}C_{4}=5, {}^{5}C_{5}=1.

Now, substitute and simplify each term:

= 1\left(\frac{32}{x^5}\right) + 5\left(\frac{16}{x^4}\right)\left(-\frac{x}{2}\right) + 10\left(\frac{8}{x^3}\right)\left(\frac{x^2}{4}\right) + 10\left(\frac{4}{x^2}\right)\left(-\frac{x^3}{8}\right) + 5\left(\frac{2}{x}\right)\left(\frac{x^4}{16}\right) + 1(1)\left(-\frac{x^5}{32}\right)

Further simplification yields:

= \frac{32}{x^5} - \frac{5 \times 16 \times x}{2 \times x^4} + \frac{10 \times 8 \times x^2}{4 \times x^3} - \frac{10 \times 4 \times x^3}{8 \times x^2} + \frac{5 \times 2 \times x^4}{16 \times x} - \frac{x^5}{32}

= \frac{32}{x^5} - \frac{40}{x^3} + \frac{20}{x} - 5x + \frac{5}{8}x^3 - \frac{x^5}{32}

The expansion is \frac{32}{x^5} - \frac{40}{x^3} + \frac{20}{x} - 5x + \frac{5}{8}x^3 - \frac{x^5}{32}.

Question 3

Expand the expression (2x - 3)^6.
Solution:

We use the Binomial Theorem to expand (2x - 3)^6. Here, \(a = 2x\), \(b = -3\), and \(n = 6\). The expansion is:

(2x-3)^6 = {}^{6}C_{0}(2x)^{6} + {}^{6}C_{1}(2x)^{5}(-3)^{1} + {}^{6}C_{2}(2x)^{4}(-3)^{2} + {}^{6}C_{3}(2x)^{3}(-3)^{3} + {}^{6}C_{4}(2x)^{2}(-3)^{4} + {}^{6}C_{5}(2x)^{1}(-3)^{5} + {}^{6}C_{6}(2x)^{0}(-3)^{6}

Calculate the binomial coefficients: {}^{6}C_{0}=1, {}^{6}C_{1}=6, {}^{6}C_{2}=15, {}^{6}C_{3}=20, {}^{6}C_{4}=15, {}^{6}C_{5}=6, {}^{6}C_{6}=1.

Now, substitute these values and simplify:

= 1(64x^{6}) + 6(32x^{5})(-3) + 15(16x^{4})(9) + 20(8x^{3})(-27) + 15(4x^{2})(81) + 6(2x)(-243) + 1(1)(729)

Perform the multiplications for each term:

= 64x^{6} - 576x^{5} + 2160x^{4} - 4320x^{3} + 4860x^{2} - 2916x + 729

The expansion of (2x - 3)^6 is 64x^{6}-576x^{5}+2160x^{4}-4320x^{3}+4860x^{2}-2916x+729.

Question 4

Expand the expression \left(\frac{x}{3} + \frac{1}{x}\right)^5.
Solution:

We use the Binomial Theorem to expand \left(\frac{x}{3} + \frac{1}{x}\right)^5. Here, \(a = \frac{x}{3}\), \(b = \frac{1}{x}\), and \(n = 5\). The expansion is:

\left(\frac{x}{3} + \frac{1}{x}\right)^5 = {}^{5}C_{0} \left(\frac{x}{3}\right)^5 + {}^{5}C_{1} \left(\frac{x}{3}\right)^4 \left(\frac{1}{x}\right)^1 + {}^{5}C_{2} \left(\frac{x}{3}\right)^3 \left(\frac{1}{x}\right)^2 + {}^{5}C_{3} \left(\frac{x}{3}\right)^2 \left(\frac{1}{x}\right)^3 + {}^{5}C_{4} \left(\frac{x}{3}\right)^1 \left(\frac{1}{x}\right)^4 + {}^{5}C_{5} \left(\frac{x}{3}\right)^0 \left(\frac{1}{x}\right)^5

Calculate the binomial coefficients: {}^{5}C_{0}=1, {}^{5}C_{1}=5, {}^{5}C_{2}=10, {}^{5}C_{3}=10, {}^{5}C_{4}=5, {}^{5}C_{5}=1.

Substitute these values and simplify each term:

= 1 \left(\frac{x^5}{243}\right) + 5 \left(\frac{x^4}{81}\right) \left(\frac{1}{x}\right) + 10 \left(\frac{x^3}{27}\right) \left(\frac{1}{x^2}\right) + 10 \left(\frac{x^2}{9}\right) \left(\frac{1}{x^3}\right) + 5 \left(\frac{x}{3}\right) \left(\frac{1}{x^4}\right) + 1 \left(1\right) \left(\frac{1}{x^5}\right)

Simplify the terms further:

= \frac{x^5}{243} + \frac{5x^4}{81x} + \frac{10x^3}{27x^2} + \frac{10x^2}{9x^3} + \frac{5x}{3x^4} + \frac{1}{x^5}

= \frac{x^5}{243} + \frac{5x^3}{81} + \frac{10x}{27} + \frac{10}{9x} + \frac{5}{3x^3} + \frac{1}{x^5}

The expansion is \frac{x^5}{243} + \frac{5x^3}{81} + \frac{10x}{27} + \frac{10}{9x} + \frac{5}{3x^3} + \frac{1}{x^5}.

Question 5

Expand \left(x+\frac{1}{x}\right)^6.
Solution:

We use the Binomial Theorem to expand \left(x+\frac{1}{x}\right)^6. Here, \(a = x\), \(b = \frac{1}{x}\), and \(n = 6\). The expansion is:

\left(x+\frac{1}{x}\right)^6 = {}^{6}C_{0} x^{6} + {}^{6}C_{1} x^{5} \left(\frac{1}{x}\right)^{1} + {}^{6}C_{2} x^{4} \left(\frac{1}{x}\right)^{2} + {}^{6}C_{3} x^{3} \left(\frac{1}{x}\right)^{3} + {}^{6}C_{4} x^{2} \left(\frac{1}{x}\right)^{4} + {}^{6}C_{5} x^{1} \left(\frac{1}{x}\right)^{5} + {}^{6}C_{6} x^{0} \left(\frac{1}{x}\right)^{6}

Calculate the binomial coefficients: {}^{6}C_{0}=1, {}^{6}C_{1}=6, {}^{6}C_{2}=15, {}^{6}C_{3}=20, {}^{6}C_{4}=15, {}^{6}C_{5}=6, {}^{6}C_{6}=1.

Substitute these values and simplify each term:

= 1(x^6) + 6(x^5)\left(\frac{1}{x}\right) + 15(x^4)\left(\frac{1}{x^2}\right) + 20(x^3)\left(\frac{1}{x^3}\right) + 15(x^2)\left(\frac{1}{x^4}\right) + 6(x)\left(\frac{1}{x^5}\right) + 1(1)\left(\frac{1}{x^6}\right)

Simplify the terms by cancelling powers of x:

= x^6 + 6x^{5-1} + 15x^{4-2} + 20x^{3-3} + 15x^{2-4} + 6x^{1-5} + \frac{1}{x^6}

= x^6 + 6x^4 + 15x^2 + 20x^0 + 15x^{-2} + 6x^{-4} + x^{-6}

= x^6 + 6x^4 + 15x^2 + 20 + \frac{15}{x^2} + \frac{6}{x^4} + \frac{1}{x^6}

The expansion is x^6 + 6x^4 + 15x^2 + 20 + \frac{15}{x^2} + \frac{6}{x^4} + \frac{1}{x^6}.

Common mistakes

  • Errors in calculating binomial coefficients (nCr).
  • Incorrectly applying the powers to terms within the binomial.
  • Sign errors, especially when dealing with negative terms in the binomial.
  • Mistakes in simplifying the final expanded terms.

Revision tips

  • Memorize the binomial expansion formula and the formula for nCr.
  • Practice expanding different types of binomial expressions, including those with fractions and negative terms.
  • Pay close attention to the powers of each term and the binomial coefficients.
  • Double-check your calculations for arithmetic errors, especially when simplifying.

Practice MCQs

Q1. What is the general form of the Binomial Theorem for expanding (a+b)^n?

Q2. In the expansion of (1-2x)^5, what is the coefficient of the x^3 term?

Q3. Which term in the expansion of (2x - 3)^6 has the highest power of x?

Q4. What is the value of ⁶C₀ in the expansion of (2x - 3)^6?

Q5. In the expansion of (x/3 + 1/x)^5, what is the term independent of x?

Frequently asked questions

What is the Binomial Theorem?

The Binomial Theorem is a formula that provides a way to expand algebraic expressions of the form (a+b)^n, where n is a positive integer. It expresses the expansion as a sum of terms involving binomial coefficients and powers of 'a' and 'b'.

How do I use the Binomial Theorem to expand (1-2x)^5?

You apply the formula (a+b)^n = Σ (nCr) * a^(n-r) * b^r. Here, a=1, b=-2x, and n=5. You calculate each term for r=0, 1, 2, 3, 4, 5 and sum them up.

What are binomial coefficients?

Binomial coefficients, denoted as nCr or (n choose r), are the numerical coefficients that appear in the expansion of a binomial. They are calculated using the formula nCr = n! / (r! * (n-r)!).

Are these solutions suitable for exam preparation?

Yes, these NCERT Solutions are specifically designed for exam preparation. They offer clear, step-by-step explanations for each problem, helping students understand the concepts and methods required to solve binomial expansion questions.

What is the general term in the binomial expansion?

The general term (or the (r+1)th term) in the expansion of (a+b)^n is given by T(r+1) = nCr * a^(n-r) * b^r.

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