Exercise Solutions Available
Step-by-step answers for all textbook questions in this chapter
Quick Summary
This chapter covers two related but distinct topics: the Binomial Theorem (expansion of (a+b)ⁿ) and Sequences/Series (AP, GP, HP, and special sums). Both are heavily tested in board exams. The binomial theorem also appears in probability and approximation problems in higher classes.
Binomial Theorem
For any positive integer n:
(a + b)ⁿ = Σ (nCr × aⁿ⁻ʳ × bʳ) for r = 0 to n
Key Points
| Term | Expression |
|---|---|
| General term (Tᵣ₊₁) | nCr × aⁿ⁻ʳ × bʳ |
| First term | aⁿ |
| Last term | bⁿ |
| Number of terms | n + 1 |
| Middle term (n even) | T(n/2 + 1) |
| Sum of coefficients | Set a = b = 1 → 2ⁿ |
| Sum of odd-position coefficients | 2ⁿ⁻¹ |
Binomial Coefficients
(a + b)¹ = 1a + 1b (a + b)² = 1a² + 2ab + 1b² (a + b)³ = 1a³ + 3a²b + 3ab² + 1b³ Pascal's triangle rows give the coefficients.
Sequences and Series
Sequence: Ordered list of numbers following a rule. Series: Sum of the terms of a sequence.
Arithmetic Progression (AP)
Terms: a, a+d, a+2d, …
| Formula | Expression |
|---|---|
| nth term | aₙ = a + (n−1)d |
| Sum of n terms | Sₙ = n/2 × [2a + (n−1)d] = n/2 × (first + last) |
| Arithmetic mean | AM = (a+b)/2 |
Geometric Progression (GP)
Terms: a, ar, ar², …
| Formula | Expression | ||
|---|---|---|---|
| nth term | aₙ = a × rⁿ⁻¹ | ||
| Sum of n terms (r ≠ 1) | Sₙ = a(rⁿ − 1)/(r − 1) | ||
| Sum to infinity (\ | r\ | < 1) | S∞ = a/(1 − r) |
| Geometric mean | GM = √(ab) |
Harmonic Progression (HP)
Terms are reciprocals of an AP: 1/a, 1/(a+d), 1/(a+2d), …
nth term of HP = 1/(nth term of corresponding AP)
Harmonic mean: HM = 2ab/(a+b)
Relation: AM ≥ GM ≥ HM (for positive numbers)
Special Series
| Series | Sum |
|---|---|
| 1 + 2 + 3 + … + n | n(n+1)/2 |
| 1² + 2² + 3² + … + n² | n(n+1)(2n+1)/6 |
| 1³ + 2³ + 3³ + … + n³ | [n(n+1)/2]² |
Worked Examples
Example 1: Find the 5th term of (2x − 3)⁸. T₅ = T₄₊₁ = 8C4 × (2x)⁴ × (−3)⁴ = 70 × 16x⁴ × 81 = 90720x⁴
Example 2: Find the sum of the AP: 3, 7, 11, …, 99. a = 3, d = 4. n: 99 = 3 + (n−1)4 → n = 25. S₂₅ = 25/2 × (3 + 99) = 25 × 51 = 1275
Example 3: A GP has first term 2 and common ratio 3. Find S₆. S₆ = 2(3⁶ − 1)/(3 − 1) = 2 × 728/2 = 728
Example 4: Find the middle term of (x + 2y)⁶. n = 6 (even), middle term = T₄ = 6C3 × x³ × (2y)³ = 20 × 8x³y³ = 160x³y³
Important Questions
One Mark
- Find the general term of (a − 2b)¹⁰.
- What is the sum of n terms of an AP with first term 5 and common difference 3?
- Find the 10th term of the GP: 3, 6, 12, …
- State the condition for an infinite GP to have a sum.
Short Answer (3 marks)
- Find the term independent of x in the expansion of (x + 1/x)⁸.
- The 4th term of an AP is 14 and the 8th term is 26. Find the AP.
- If the sum of 3 numbers in GP is 26 and their product is 216, find the numbers.
Long Answer (5 marks)
- Prove that 1² + 2² + … + n² = n(n+1)(2n+1)/6 using the method of differences or mathematical induction.
- Find the sum to n terms: 1×2 + 2×3 + 3×4 + … + n(n+1).
Common Mistakes
- Binomial: writing Tᵣ instead of Tᵣ₊₁ for the general term — remember the (r+1)th term formula.
- Forgetting to account for the negative sign in (a − b)ⁿ when writing the general term.
- AP: using Sₙ = n/2 × (2a + nd) instead of n/2 × (2a + (n−1)d).
- GP: using the wrong formula for sum when r < 0 — the formula works, just be careful with signs.
- Confusing AM, GM, HM — AM ≥ GM ≥ HM for positive numbers only.
Board Exam Tips
- "Find the term independent of x" is a standard 3-mark binomial question — set the power of x to zero in the general term and solve for r.
- Special series sums (Σn, Σn², Σn³) appear in 5-mark problems — memorise all three.
- AP and GP sum problems where conditions are given (e.g., sum of first n terms, ratio of sums) appear every year.
- JEE: Binomial coefficients and properties, and sum of series involving GP, are common.