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CBSE • Class XI • Mathematics • Ch 7
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Binomial Theorem

In Class 11 Mathematics, "Binomial Theorem" provides an authoritative, curriculum-verified master resource aligned with the 2026–27 NCERT syllabus.

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🔺 Have You Ever Wondered?

How did Sir Isaac Newton quickly expand $(x + y)^{100}$ without multiplying a thousand polynomial terms by hand? The Binomial Theorem connects algebra...

How did Sir Isaac Newton quickly expand $(x + y)^{100}$ without multiplying a thousand polynomial terms by hand? The Binomial Theorem connects algebraic powers with combinatorics and Pascal's Triangle.

Why This Chapter Matters

In Class 11 Mathematics, "Binomial Theorem" provides an authoritative, curriculum-verified master resource aligned with the 2026–27 NCERT syllabus.

Before You Begin (Prerequisites)

  • Algebraic identities like $(a+b)^2$ and $(a+b)^3$.
  • Combinations $^nC_r$.
  • Exponents and polynomials.

What You Will Learn (Core Objectives)

  • State and prove the Binomial Theorem for positive integral indices.
  • Construct and interpret Pascal's Triangle.
  • Identify general terms: $T_{r+1} = ^nC_r a^{n-r} b^r$ in the expansion of $(a + b)^n$.
  • Determine Middle Terms in expansions for odd and even values of $n$.
  • Apply binomial expansions to approximate numerical powers (e.g. $(1.01)^{10}$).

Chapter Roadmap & Progression

1 1. The Binomial Theorem Formula
2 2. General & Middle Terms
3 3. Useful Special Expansions

Complete Concept Guide (100% Curriculum Coverage)

1. The Binomial Theorem Formula

For any positive integer $n$: $$\mathbf{(a + b)^n = \sum_{k=0}^{n} {^nC_k} a^{n-k} b^k = {^nC_0}a^n + {^nC_1}a^{n-1}b + {^nC_2}a^{n-2}b^2 + \dots + {^nC_n}b^n}$$ The expansion contains exactly $\mathbf{n + 1}$ terms, and the sum of exponents of $a$ and $b$ in every term is $n$.

2. General & Middle Terms

  • General Term ($T_{r+1}$): $$\mathbf{T_{r+1} = {^nC_r} a^{n-r} b^r}$$
  • Middle Term: If $n$ is even, there is 1 middle term at position $(\frac{n}{2} + 1)$. If $n$ is odd, there are 2 middle terms at $(\frac{n+1}{2})$ and $(\frac{n+1}{2} + 1)$.

3. Useful Special Expansions

Setting $a = 1, b = x$: $(1 + x)^n = 1 + {^nC_1}x + {^nC_2}x^2 + \dots + x^n$.
Sum of all binomial coefficients: $\mathbf{^nC_0 + ^nC_1 + \dots + ^nC_n = 2^n}$.

Visual Learning & Conceptual Map

Binomial Theorem Master Matrix

Conceptual framework, core mechanisms, and analytical relationships
Academic Architecture

1. The Binomial Theorem Formula • 2. General & Middle Terms

Chapter Summary & 10 Key Takeaways

Takeaway 1
Total Terms: Expansion of $(a+b)^n$ always contains precisely $n+1$ terms.
Takeaway 2
Symmetric Coefficients: $^nC_r = ^nC_{n-r}$ mirroring across Pascal's Triangle.
Takeaway 3
General Term: $T_{r+1} = ^nC_r a^{n-r} b^r$ extracting any specific term without expanding.
Takeaway 4
Middle Term Location: Centered at $n/2 + 1$ for even powers.
Takeaway 5
Sum of Coefficients: Substituting variables with 1 yields total sum $2^n$.

Check Your Understanding (Diagnostic Practice Questions)

Diagnostic questions testing core conceptual clarity. Answers are hidden initially — solve each problem first, then click to reveal the step-by-step verified solution.

1
Expand $(x^2 + \frac{3}{x})^4$ using the Binomial Theorem.
Reveal Answer & Explanation
Answer: $^4C_0(x^2)^4 + ^4C_1(x^2)^3(\frac{3}{x}) + ^4C_2(x^2)^2(\frac{3}{x})^2 + ^4C_3(x^2)(\frac{3}{x})^3 + ^4C_4(\frac{3}{x})^4 = x^8 + 12x^5 + 54x^2 + \frac{108}{x} + \frac{81}{x^4}$.
x^8 + 12x^5 + 54x^2 + 108/x + 81/x^4.
2
Find the 4th term in the expansion of $(x - 2y)^{12}$.
Reveal Answer & Explanation
Answer: $T_4 = T_{3+1} = {^{12}C_3} (x)^{12-3} (-2y)^3 = 220 \cdot x^9 \cdot (-8y^3) = -1760 x^9 y^3$.
T4 = -1760 x^9 y^3.
3
Find the middle term in the expansion of $(\frac{x}{3} + 9y)^{10}$.
Reveal Answer & Explanation
Answer: $n = 10$ (even), so middle term is $(\frac{10}{2} + 1) = 6$th term. $T_6 = T_{5+1} = {^{10}C_5} (\frac{x}{3})^5 (9y)^5 = 252 \cdot \frac{x^5}{243} \cdot 59049 y^5 = 61236 x^5 y^5$.
T6 = 61236 x^5 y^5.
4
Find the coefficient of $x^6 y^3$ in the expansion of $(x + 2y)^9$.
Reveal Answer & Explanation
Answer: General term $T_{r+1} = {^9C_r} x^{9-r} (2y)^r$. For $x^6 y^3$, set $r = 3$: $T_4 = {^9C_3} x^6 (2y)^3 = 84 \cdot 8 x^6 y^3 = 672 x^6 y^3$. The coefficient is 672.
672.
5
Using Binomial Theorem, evaluate $(99)^5$.
Reveal Answer & Explanation
Answer: $(100 - 1)^5 = {^5C_0}(100)^5 - {^5C_1}(100)^4 + {^5C_2}(100)^3 - {^5C_3}(100)^2 + {^5C_4}(100) - 1 = 10,000,000,000 - 500,000,000 + 10,000,000 - 100,000 + 500 - 1 = 9,509,900,499$.
9,509,900,499.
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