1. Chapter at a Glance
You know (a + b)² = a² + 2ab + b² and (a + b)³ = a³ + 3a²b + 3ab² + b³. But what about (a + b)¹⁰ or (98)⁵? Multiplying out is tedious. The binomial theorem gives the expansion of (a + b)ⁿ directly, with coefficients taken from Pascal's triangle — a number pattern known in India much earlier as Meru Prastāra, described in connection with Piṅgala's work on poetic metres and explained by Halāyudha (10th century). This chapter states and proves the theorem for positive integer n and applies it to expansions and approximations.
- Pascal's triangle (Meru Prastāra) and its rule.
- Statement of the binomial theorem for positive integral index.
- Proof by the counting (combinatorial) argument.
- Special cases: (x − y)ⁿ, (1 + x)ⁿ, (1 − x)ⁿ.
- Applications: expansions, large powers, comparing numbers, divisibility.
- Enrichment: the general term.
2. Pascal's Triangle (Meru Prastāra)
Write the coefficients of (a + b)ⁿ for n = 0, 1, 2, …. Each row starts and ends with 1, and every inner number is the sum of the two numbers just above it.
Row n lists ⁿC₀, ⁿC₁, …, ⁿCₙ. The sum of row n is 2ⁿ.
3. The Binomial Theorem
In sigma notation: (a + b)ⁿ = Σ (k = 0 to n) ⁿCₖ aⁿ⁻ᵏ bᵏ.
- There are n + 1 terms.
- Powers of a decrease from n to 0; powers of b increase from 0 to n.
- In each term, the powers of a and b add up to n.
- Coefficients equidistant from the ends are equal (ⁿCᵣ = ⁿCₙ₋ᵣ).
Why it is true (counting proof): (a + b)ⁿ = (a + b)(a + b)…(a + b), n factors. A term aⁿ⁻ᵏbᵏ arises by picking b from exactly k of the n brackets (and a from the rest). This can be done in ⁿCₖ ways, so ⁿCₖ is its coefficient. (NCERT also proves it by mathematical induction.)
4. Special Cases
| Expression | Expansion |
|---|---|
| (x − y)ⁿ | ⁿC₀xⁿ − ⁿC₁xⁿ⁻¹y + ⁿC₂xⁿ⁻²y² − … + (−1)ⁿ ⁿCₙyⁿ |
| (1 + x)ⁿ | ⁿC₀ + ⁿC₁x + ⁿC₂x² + … + ⁿCₙxⁿ |
| (1 − x)ⁿ | ⁿC₀ − ⁿC₁x + ⁿC₂x² − … + (−1)ⁿ ⁿCₙxⁿ |
Putting x = 1 in (1 + x)ⁿ: ⁿC₀ + ⁿC₁ + … + ⁿCₙ = 2ⁿ. Putting x = 1 in (1 − x)ⁿ: alternating sum = 0.
5. Expansions and Applications
Expand (x + 2/x)⁴.
Solution: x⁴ + 4x³(2/x) + 6x²(4/x²) + 4x(8/x³) + 16/x⁴ = x⁴ + 8x² + 24 + 32/x² + 16/x⁴.
Compute (98)⁵.
Solution: (100 − 2)⁵ = 100⁵ − 5(100⁴)(2) + 10(100³)(4) − 10(100²)(8) + 5(100)(16) − 32 = 10000000000 − 1000000000 + 40000000 − 800000 + 8000 − 32 = 9039207968.
Which is larger: (1.01)¹⁰⁰⁰⁰⁰⁰ or 10,000?
Solution: (1 + 0.01)¹⁰⁰⁰⁰⁰⁰ = 1 + 1000000 × 0.01 + (positive terms) = 1 + 10000 + … > 10000. So (1.01)¹⁰⁰⁰⁰⁰⁰ is larger.
Show that 6ⁿ − 5n leaves remainder 1 when divided by 25.
Solution: 6ⁿ = (1 + 5)ⁿ = 1 + 5n + ⁿC₂5² + ⁿC₃5³ + … So 6ⁿ − 5n = 1 + 25(ⁿC₂ + ⁿC₃5 + …) — remainder 1. ∎
6. Enrichment: The General Term
The (k + 1)th term of (a + b)ⁿ is Tₖ₊₁ = ⁿCₖ aⁿ⁻ᵏ bᵏ. It is handy for finding a particular term or coefficient without the full expansion.
Find the coefficient of x⁵ in (x + 3)⁸.
Solution: Tₖ₊₁ = ⁸Cₖ x⁸⁻ᵏ 3ᵏ. 8 − k = 5 → k = 3. Coefficient = ⁸C₃ × 27 = 56 × 27 = 1512.
7. Solved Problems (Board Pattern)
Expand (1 − 2x)⁵.
Solution: 1 − 5(2x) + 10(4x²) − 10(8x³) + 5(16x⁴) − 32x⁵ = 1 − 10x + 40x² − 80x³ + 80x⁴ − 32x⁵.
Find (a + b)⁴ − (a − b)⁴ and hence evaluate (√3 + √2)⁴ − (√3 − √2)⁴.
Solution: Odd-position terms cancel: 2[4a³b + 4ab³] = 8ab(a² + b²). With a = √3, b = √2: 8√6 × 5 = 40√6.
Using the binomial theorem, evaluate (102)⁵.
Solution: (100 + 2)⁵ = 10¹⁰ + 10⁹ + 4×10⁷ + 8×10⁵ + 8×10³ + 32 = 11040808032.
8. Master Formula Table
| Result | Formula |
|---|---|
| Binomial theorem | (a + b)ⁿ = Σ ⁿCₖ aⁿ⁻ᵏ bᵏ |
| Number of terms | n + 1 |
| General term (enrichment) | Tₖ₊₁ = ⁿCₖ aⁿ⁻ᵏ bᵏ |
| Sum of coefficients | ΣⁿCₖ = 2ⁿ |
| Alternating sum | ⁿC₀ − ⁿC₁ + ⁿC₂ − … = 0 |
| Pascal's rule | ⁿCᵣ + ⁿCᵣ₋₁ = ⁿ⁺¹Cᵣ |
| (a + b)ⁿ + (a − b)ⁿ | 2[terms with even powers of b] |
| (a + b)ⁿ − (a − b)ⁿ | 2[terms with odd powers of b] |
9. Master Summary Table
| Topic | Key idea | Exam focus |
|---|---|---|
| Pascal's triangle | Add the two above | Write rows; coefficients |
| Theorem | ⁿCₖ = ways to pick k b's | Statement, expansions |
| Negative second term | Alternating signs | (x − y)ⁿ |
| Large powers | Write as (100 ± k)ⁿ | (98)⁵, (101)⁴ |
| Comparisons, divisibility | (1 + x)ⁿ ≥ 1 + nx | Prove remainders |
Chart — sum of row n of Pascal's triangle (= 2ⁿ):
10. Mind Map
- Add above
- Row sum 2ⁿ
- Σ ⁿCₖ aⁿ⁻ᵏbᵏ
- n + 1 terms
- (x − y)ⁿ
- (1 ± x)ⁿ
- Big powers
- Remainders, comparison
11. Tips and Tricks
- For (a − b)ⁿ, expand as (a + (−b))ⁿ — the signs alternate automatically.
- For powers of numbers near 100, use (100 ± k)ⁿ.
- For remainder problems, write the base as (multiple of d) ± 1.
- Forgetting to raise the coefficient: (2x)³ = 8x³, not 2x³.
- Miscounting terms: (a + b)⁷ has 8 terms.
- Sign errors with (−b)ᵏ for odd k.
12. Important One-liners
- (a + b)ⁿ has n + 1 terms.
- Coefficients equidistant from the ends are equal.
- Sum of binomial coefficients = 2ⁿ.
- Pascal's triangle was known in India as Meru Prastāra.
- ⁿCₖ is the coefficient of aⁿ⁻ᵏbᵏ.
13. Exam-pattern Questions with Detailed Answers
The number of terms in the expansion of (2x + 3y)⁹ is:
- 9
- 10
- 8
- 11
Show answer
n + 1 = 10.
Answer: (B)
Write row 6 of Pascal's triangle.
Show answer
1, 6, 15, 20, 15, 6, 1 (sum 64).
Answer: 1 6 15 20 15 6 1
Expand (1 + x)⁴.
Show answer
1 + 4x + 6x² + 4x³ + x⁴.
Answer: 1 + 4x + 6x² + 4x³ + x⁴
Expand (2x − 3)⁶.
Show answer
64x⁶ − 6(32x⁵)(3) + 15(16x⁴)(9) − 20(8x³)(27) + 15(4x²)(81) − 6(2x)(243) + 729 = 64x⁶ − 576x⁵ + 2160x⁴ − 4320x³ + 4860x² − 2916x + 729.
Answer: As above
Evaluate (101)⁴ using the binomial theorem.
Show answer
(100 + 1)⁴ = 10⁸ + 4×10⁶ + 6×10⁴ + 4×10² + 1 = 104060401.
Answer: 104060401
Expand (x/3 + 1/x)⁵.
Show answer
x⁵/243 + 5x³/81 + 10x/27 + 10/(9x) + 5/(3x³) + 1/x⁵.
Answer: As above
Find (x + 1)⁶ + (x − 1)⁶ and hence (√2 + 1)⁶ + (√2 − 1)⁶.
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2[x⁶ + 15x⁴ + 15x² + 1]. x = √2: 2[8 + 60 + 30 + 1] = 198.
Answer: 198
Show that 9ⁿ⁺¹ − 8n − 9 is divisible by 64 for every positive integer n.
Show answer
9ⁿ⁺¹ = (1 + 8)ⁿ⁺¹ = 1 + 8(n + 1) + ⁿ⁺¹C₂ 8² + … So 9ⁿ⁺¹ − 8n − 9 = 64(ⁿ⁺¹C₂ + ⁿ⁺¹C₃ 8 + …). ∎
Answer: Proved
Find the approximate value of (0.99)⁵ using the first three terms.
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(1 − 0.01)⁵ ≈ 1 − 0.05 + 10(0.0001) = 0.951.
Answer: ≈ 0.951
Find the coefficient of x⁶y³ in (x + 2y)⁹.
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k = 3: ⁹C₃ × 2³ = 84 × 8 = 672.
Answer: 672
Using the general term, find the term independent of x in (x² + 1/x)⁹.
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Tₖ₊₁ = ⁹Cₖ x^(2(9 − k)) x^(−k) = ⁹Cₖ x^(18 − 3k). 18 − 3k = 0 → k = 6. Term = ⁹C₆ = 84.
Answer: 84
Prove that Σ (k = 0 to n) 3ᵏ ⁿCₖ = 4ⁿ.
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(1 + 3)ⁿ = Σ ⁿCₖ 1ⁿ⁻ᵏ 3ᵏ = 4ⁿ. ∎
Answer: Proved
14. Quick Revision Summary
- (a + b)ⁿ = Σ ⁿCₖ aⁿ⁻ᵏbᵏ; n + 1 terms; powers sum to n.
- Coefficients from Pascal's triangle (Meru Prastāra); row sum 2ⁿ.
- (a − b)ⁿ: alternate signs.
- Use (100 ± k)ⁿ for big powers and (1 + d)ⁿ for remainders.
- General term Tₖ₊₁ = ⁿCₖ aⁿ⁻ᵏbᵏ finds specific coefficients.