By the end of this chapter you'll be able to…

  • 1Construct Pascal's Triangle and state its relationship to binomial coefficients nCr
  • 2State and apply the Binomial Theorem to expand (a+b)ⁿ for positive integer n
  • 3Use the general term formula Tᵣ₊₁ = nCᵣ·aⁿ⁻ʳ·bʳ to find any specified term in an expansion
  • 4Identify the middle term(s) of a binomial expansion for both even and odd n
  • 5Apply special cases (1+x)ⁿ and (1−x)ⁿ and use the sum-of-coefficients = 2ⁿ property
💡
Why this chapter matters
The Binomial Theorem provides a systematic formula for expanding (a+b)ⁿ for any positive integer n, avoiding repeated multiplication. The general term formula is the key tool for finding any specific term in an expansion — a type of question that appears in virtually every JEE Main paper.

Before you start — revise these

A 5-minute refresher here will save you 30 minutes of confusion below.

Binomial Theorem

"Algebra begins with (a + b)². The Binomial Theorem is what happens when the 2 becomes n."

1. Chapter Overview

Expanding (a + b)² = a² + 2ab + b² is easy. But what about (a + b)¹⁰? (a + b)ⁿ? The Binomial Theorem provides the formula: the expansion uses BINOMIAL COEFFICIENTS (nCr) and PATTERNS in the powers. This chapter covers: the theorem, Pascal's triangle, general and middle terms.


2. Pascal's Triangle

A triangular array where each number is the SUM of the two numbers above it:

        1          ← (a + b)⁰
       1 1         ← (a + b)¹
      1 2 1        ← (a + b)²
     1 3 3 1       ← (a + b)³
    1 4 6 4 1      ← (a + b)⁴
   1 5 10 10 5 1   ← (a + b)⁵
  • Row n (starting from 0) gives the coefficients of (a + b)ⁿ
  • Each number = nCr. Row 4: ⁴C₀=1, ⁴C₁=4, ⁴C₂=6, ⁴C₃=4, ⁴C₄=1

3. The Binomial Theorem

For any positive integer n:

Expanded:

Observations

  • There are (n + 1) terms in the expansion
  • In each term: power of a DECREASES from n to 0. Power of b INCREASES from 0 to n.
  • Sum of powers of a and b in each term = n
  • The coefficients are SYMMETRIC: nCr = nC(n-r)
  • nC₀ = nCn = 1

4. General Term

The (r + 1)th term (counting from r = 0):

Middle Term(s)

  • If n is EVEN: there is ONE middle term. Term number = (n/2 + 1). r = n/2.
  • If n is ODD: there are TWO middle terms. Terms at r = (n-1)/2 and r = (n+1)/2.

5. Special Cases

(1 + x)ⁿ

(1 — x)ⁿ

Same expansion but with ALTERNATING SIGNS:


6. Properties of Binomial Coefficients

  • Sum of all coefficients: nC₀ + nC₁ + nC₂ + ... + nCn = 2ⁿ
  • Sum of coefficients at EVEN positions = sum at ODD positions = 2ⁿ⁻¹
  • nCr + nC(r-1) = (n+1)Cr

7. Exam Focus

  1. Pascal's triangle — how constructed, relationship to nCr
  2. Binomial theorem formula — expansion of (a + b)ⁿ
  3. General term Tr+1 = nCr · a^(n-r) · b^r
  4. Middle term(s) — formula for even and odd n
  5. Expansion of (1 + x)ⁿ and (1 — x)ⁿ

8. Key Formulas

  • (a + b)ⁿ = Σ nCr a^(n-r) b^r (r = 0 to n)
  • General term: Tr+1 = nCr a^(n-r) b^r
  • Sum of coefficients = 2ⁿ

9. Conclusion

The Binomial Theorem is a PATTERN-MACHINE:

  • PASCAL'S TRIANGLE: The coefficients for small n — visible at a glance
  • THE THEOREM: For ANY n — a systematic expansion using nCr
  • GENERAL TERM: Find ANY term without writing the whole expansion

'The binomial theorem is a thing of beauty — a simple pattern that generates an infinity of terms from two numbers and a power.'

Key formulas & results

Everything you need to memorise, in one card. Screenshot this for revision.

Binomial Theorem
(a+b)ⁿ = Σ(r=0 to n) nCr · aⁿ⁻ʳ · bʳ = nC₀aⁿ + nC₁aⁿ⁻¹b + nC₂aⁿ⁻²b² + ... + nCₙbⁿ
(n+1) terms total; powers of a decrease, powers of b increase, sum of powers in each term = n
General Term (r+1)th term
Tᵣ₊₁ = nCᵣ · aⁿ⁻ʳ · bʳ
r starts at 0, so the 1st term is T₁ (r=0), the 2nd is T₂ (r=1), etc.
Middle Term — n even
If n is even: one middle term = T(n/2+1) at r = n/2
Expansion of (a+b)ⁿ has n+1 terms; the (n/2+1)th is the single middle term
Middle Terms — n odd
If n is odd: two middle terms = T((n+1)/2) at r=(n−1)/2 and T((n+3)/2) at r=(n+1)/2
For odd n, n+1 is even so there are exactly two central terms
Sum of Binomial Coefficients
nC₀ + nC₁ + nC₂ + ... + nCₙ = 2ⁿ
Obtained by substituting a=1, b=1 in (a+b)ⁿ
Pascal's Identity
nCᵣ + nCᵣ₋₁ = (n+1)Cᵣ
Each entry in Pascal's Triangle = sum of the two entries directly above it
⚠️

Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT
Starting the general term count from r=1 instead of r=0
The general term is Tᵣ₊₁. At r=0, T₁ is the FIRST term. At r=1, T₂ is the SECOND term. Always start r at 0.
WATCH OUT
Substituting the wrong exponents in Tᵣ₊₁ = nCᵣ·aⁿ⁻ʳ·bʳ
In (a+b)ⁿ: power of a = n−r, power of b = r. They must always sum to n. Double-check: (n−r)+r=n.
WATCH OUT
Forgetting to account for coefficients of a and b when they are not 1
In (2x + 3y)⁴, the 'a' is 2x and the 'b' is 3y. So T₃ = ⁴C₂·(2x)²·(3y)² — raise the ENTIRE term (coefficient and variable together) to the power.
WATCH OUT
Saying there is one middle term for any n
For even n: ONE middle term. For odd n: TWO middle terms. The expansion has n+1 terms — if n+1 is odd (n even), there is one middle term; if n+1 is even (n odd), there are two.

Practice problems

Try each one yourself before tapping "Show solution". Active recall > rereading.

Q1EASY· Expansion
Expand (x + 2)⁴ using the Binomial Theorem.
Show solution
(x+2)⁴ = ⁴C₀x⁴ + ⁴C₁x³(2) + ⁴C₂x²(2²) + ⁴C₃x(2³) + ⁴C₄(2⁴) = x⁴ + 4×2x³ + 6×4x² + 4×8x + 16 = x⁴ + 8x³ + 24x² + 32x + 16.
Q2MEDIUM· General Term
Find the 5th term in the expansion of (2x − 3y)⁸.
Show solution
T₅ = T(4+1): r=4. T₅ = ⁸C₄·(2x)⁸⁻⁴·(−3y)⁴ = 70·(2x)⁴·(−3y)⁴ = 70·16x⁴·81y⁴ = 70×16×81×x⁴y⁴ = 90720x⁴y⁴.
Q3HARD· Middle Term
Find the middle term(s) in the expansion of (x/2 + 3/x)⁸.
Show solution
n=8 (even), so ONE middle term: T₅ (at r=4). T₅ = ⁸C₄·(x/2)⁴·(3/x)⁴ = 70·(x⁴/16)·(81/x⁴) = 70·81/16 = 5670/16 = 2835/8.

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

  • (a+b)ⁿ has exactly (n+1) terms in its expansion
  • General term: Tᵣ₊₁ = nCᵣ·aⁿ⁻ʳ·bʳ where r = 0, 1, 2, ..., n
  • Coefficients are symmetric: nCᵣ = nCₙ₋ᵣ, so first and last coefficients are equal (=1)
  • Sum of all binomial coefficients = 2ⁿ (put a=b=1)
  • Pascal's Identity: nCᵣ + nCᵣ₋₁ = (n+1)Cᵣ — each Pascal's Triangle entry = sum of two above
  • Middle term: one middle term if n is even (T(n/2+1)); two middle terms if n is odd
  • For (1+x)ⁿ: coefficients are 1, n, n(n-1)/2, ... — directly from Pascal's Triangle row n
  • For (1−x)ⁿ: same coefficients but alternating signs: +,−,+,−,...

CBSE marks blueprint

Where the marks come from in this chapter — so you can plan your prep.

Typical chapter weightage: 6-8 marks

Question typeMarks eachTypical countWhat it tests
Short Answer21Expanding small powers, sum of coefficients
Long Answer4-61Finding general term, middle term, coefficient of specific power
Prep strategy
  • Memorise the general term formula Tᵣ₊₁ = nCᵣ·aⁿ⁻ʳ·bʳ and practise applying it with different values of a, b, and n
  • For 'find the term independent of x' questions: write the general term, set the power of x to zero, and solve for r
  • Verify middle term calculations by confirming the powers of a and b in the term add up to n

Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

Probability Distributions

The Binomial probability distribution P(X=r) = nCᵣ·pʳ·(1−p)ⁿ⁻ʳ directly uses the Binomial Theorem — giving the probability of r successes in n independent trials.

Approximation in Physics and Engineering

When x is very small, (1+x)ⁿ ≈ 1+nx (first two terms of the expansion) — used to approximate expressions in mechanics, optics, and error analysis.

Computer Science — Combinatorial Algorithms

The number of subsets of a set of size n is 2ⁿ (sum of all nCᵣ) — the time complexity of certain brute-force algorithms in computer science.

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

  1. For every binomial expansion question, write the general term Tᵣ₊₁ = nCᵣ·aⁿ⁻ʳ·bʳ first — it solves ALL sub-types
  2. When a or b has a coefficient (e.g., (2x)ⁿ⁻ʳ), expand the power fully: (2x)⁴ = 16x⁴, not 2x⁴
  3. For middle term questions: state n, determine if n is even or odd, then compute the appropriate term
  4. Check your general term by verifying that powers of a and b sum to n

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

  • Binomial coefficients mod prime: Lucas' theorem gives nCr mod p efficiently — important in number theory competitions
  • Multinomial theorem: extension to (a+b+c)ⁿ = Σ n!/(p!q!r!) · aᵖbᵍcʳ where p+q+r=n

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

CBSE Class 11 BoardHigh
JEE MainVery High
JEE AdvancedHigh

Questions students ask

The real ones — pulled from the Q&A community and tutor sessions.

Write the general term Tᵣ₊₁, collect the total power of x, set it equal to the required power, solve for r, then substitute back into Tᵣ₊₁.

It means the power of x in that term is 0 (x⁰=1). Set the x-exponent in the general term equal to 0, solve for r, then compute Tᵣ₊₁.

Substitute x=1 (and a=b=1) in (a+b)ⁿ = Σ nCᵣ·aⁿ⁻ʳ·bʳ: (1+1)ⁿ = Σ nCᵣ, so 2ⁿ = nC₀+nC₁+...+nCₙ.
Verified by the tuition.in editorial team
Last reviewed on 26 May 2026. Written and reviewed by subject-matter experts — read about our process.
Editorial process →
Header Logo