What this chapter is about
The Binomial Theorem provides a systematic method to expand expressions of the form (a + b)ⁿ, where n is a positive integer. Before this theorem, expanding (a + b)² or (a + b)³ requires repeated multiplication, which becomes impractical for higher powers. This chapter gives you a formula that directly writes out any term of such an expansion without multiplying step by step.
A Class 11 student meets this topic now because it builds on the concepts of permutations and combinations studied earlier. The coefficients appearing in binomial expansions are exactly the combinations ⁿCᵣ, connecting algebra with counting principles. This linkage deepens your understanding of both areas.
After studying this chapter, you should be able to expand any binomial raised to a positive integer power, find any specific term in the expansion, identify the middle term or terms, and apply these ideas to problems involving approximations and coefficient calculations.
Key ideas
- The expansion of (a + b)ⁿ has exactly (n + 1) terms, starting from a term with aⁿ and ending with a term containing bⁿ.
- Each term in the expansion has the form ⁿCᵣ × aⁿ⁻ʳ × bʳ, where r takes values 0, 1, 2, …, n.
- The coefficients ⁿC₀, ⁿC₁, ⁿC₂, …, ⁿCₙ are called binomial coefficients and appear in Pascal's Triangle, where each entry equals the sum of the two entries directly above it.
- The sum of all binomial coefficients in (1 + x)ⁿ equals 2ⁿ, obtained by putting x = 1.
- In the expansion of (a + b)ⁿ, the (r + 1)th term is called the general term, written as Tᵣ₊₁ = ⁿCᵣ × aⁿ⁻ʳ × bʳ.
- If n is even, there is one middle term at position (n/2 + 1). If n is odd, there are two middle terms at positions (n + 1)/2 and (n + 3)/2.
- Binomial coefficients satisfy the property ⁿCᵣ = ⁿCₙ₋ᵣ, which explains why coefficients are symmetric about the middle of the expansion.
Formulas and facts to remember
Binomial expansion: (a + b)ⁿ = ⁿC₀ aⁿ + ⁿC₁ aⁿ⁻¹ b + ⁿC₂ aⁿ⁻² b² + … + ⁿCₙ bⁿ — the complete expansion for any positive integer n.
General term: Tᵣ₊₁ = ⁿCᵣ × aⁿ⁻ʳ × bʳ — gives the (r + 1)th term directly without expanding everything.
Binomial coefficient formula: ⁿCᵣ = n! / [r! × (n − r)!] — the number of ways to choose r objects from n.
Sum of coefficients: ⁿC₀ + ⁿC₁ + ⁿC₂ + … + ⁿCₙ = 2ⁿ — put a = 1, b = 1 in the expansion.
Alternating sum: ⁿC₀ − ⁿC₁ + ⁿC₂ − ⁿC₃ + … = 0 — put a = 1, b = −1 in the expansion.
Middle term (n even): The middle term is T₍ₙ/₂₎₊₁.
Middle terms (n odd): The two middle terms are T₍ₙ₊₁₎/₂ and T₍ₙ₊₃₎/₂.
Worked examples
Example 1: Expand (2x + 3)⁴ using the Binomial Theorem.
Here a = 2x, b = 3, n = 4.
The expansion is: ⁴C₀ (2x)⁴ + ⁴C₁ (2x)³ (3) + ⁴C₂ (2x)² (3)² + ⁴C₃ (2x)(3)³ + ⁴C₄ (3)⁴
Calculate each term:
- ⁴C₀ (2x)⁴ = 1 × 16x⁴ = 16x⁴
- ⁴C₁ (2x)³ (3) = 4 × 8x³ × 3 = 96x³
- ⁴C₂ (2x)² (9) = 6 × 4x² × 9 = 216x²
- ⁴C₃ (2x)(27) = 4 × 2x × 27 = 216x
- ⁴C₄ (81) = 1 × 81 = 81
Therefore, (2x + 3)⁴ = 16x⁴ + 96x³ + 216x² + 216x + 81.
Example 2: Find the 5th term in the expansion of (x − 2y)⁷.
Here a = x, b = −2y, n = 7. The 5th term means r + 1 = 5, so r = 4.
T₅ = ⁷C₄ × x⁷⁻⁴ × (−2y)⁴
Calculate: ⁷C₄ = 7!/(4! × 3!) = (7 × 6 × 5)/(3 × 2 × 1) = 35
(−2y)⁴ = 16y⁴ (positive because the power is even)
T₅ = 35 × x³ × 16y⁴ = 560x³y⁴.
Example 3: Find the coefficient of x⁵ in the expansion of (1 + x)⁸.
Using the general term Tᵣ₊₁ = ⁸Cᵣ × 1⁸⁻ʳ × xʳ = ⁸Cᵣ × xʳ.
For the term containing x⁵, we need r = 5.
Coefficient = ⁸C₅ = 8!/(5! × 3!) = (8 × 7 × 6)/(3 × 2 × 1) = 56.
The coefficient of x⁵ is 56.
Common mistakes
Forgetting that (r + 1)th term uses r in the formula → always set r + 1 equal to the term number you want, then solve for r.
Ignoring the negative sign in expressions like (a − b)ⁿ → substitute b as (−b) and track the sign through the power.
Miscounting the number of terms as n instead of (n + 1) → remember that r goes from 0 to n, giving (n + 1) values.
Confusing ⁿCᵣ with ⁿPᵣ → binomial coefficients use combinations (order does not matter), not permutations.
Errors in finding middle terms when n is odd → recognise that odd n gives two middle terms, not one.
Quick revision
- (a + b)ⁿ expands into (n + 1) terms with coefficients from Pascal's Triangle.
- General term: Tᵣ₊₁ = ⁿCᵣ × aⁿ⁻ʳ × bʳ.
- Sum of all binomial coefficients equals 2ⁿ.
- ⁿCᵣ = ⁿCₙ₋ᵣ makes coefficients symmetric.
- Even n has one middle term; odd n has two.