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Binomial Theorem, Sequences and Series — Exercise Solutions

1. How to Use This Page

This page covers all exercise types from the TN Class 11 Maths Chapter 5 textbook. Work through each section in order, or jump to a specific exercise using the links below. Every answer includes a worked explanation — not just the final result.

Quick navigation:

1. [Key Concepts (Quick Recap)](#2-key-concepts-quick-recap) 2. [I. Choose the Best Answer (MCQs)](#i-choose-the-best-answer-mcqs) 3. [II. Fill in the Blanks](#ii-fill-in-the-blanks) 4. [III. True or False](#iii-true-or-false-with-corrections) 5. [IV. Match the Following]((#iv-match-the-following)) 6. [V. Answer Briefly (2-mark)](#v-answer-briefly-2-mark-questions) 7. [VI. Answer in Detail (5-mark)](#vi-answer-in-detail-5-mark-questions) 8. [Common Mistakes to Avoid](#8-common-mistakes-to-avoid) 9. [Exam Checklist](#9-exam-checklist)

2. Key Concepts (Quick Recap)

2.1 Binomial Theorem

For any positive integer n:

(a + b)ⁿ = Σ (r=0 to n) nCr · a^(n−r) · b^r

2.2 Sequences

2.3 Arithmetic Progression (AP)

2.4 Geometric Progression (GP)

2.5 Harmonic Progression (HP)

2.6 AM, GM, HM Relations

2.7 Special Series Sums

I. Choose the Best Answer (MCQs)

Q1. The 5th term in the expansion of (x + 2)⁷ is

Using T(r+1) = nCr · a^(n−r) · b^r, for the 5th term r = 4:

Q2. If the sum of first n terms of an AP is Sₙ = 3n² + 5n, then the common difference is

Q3. The sum of the infinite series 8 + 4 + 2 + … is

This is an infinite GP with a = 8, r = 1/2.

Q4. If AM = 10 and HM = 9 for two positive numbers, then GM is

Q5. Σn³ / Σn (from 1 to n) equals

Q6. The coefficient of x⁵ in the expansion of (1 + x)¹⁰ is

II. Fill in the Blanks

1. The middle term in the expansion of (x + y)⁸ is ______.

2. If a, b, c are in AP, then 2b = ______.

3. The sum 1 + 2 + 3 + … + 100 = ______.

4. The common ratio of the GP 3, −6, 12, −24, … is ______.

5. In the expansion of (a + b)ⁿ, the sum of all binomial coefficients is ______.

6. The 4th term of the HP 1/3, 1/5, 1/7, … is ______.

7. If Σn = 210 for n terms, then n = ______.

8. For the GP 5, 10, 20, …, the 6th term is ______.

9. If GM = 6 and HM = 4 for two numbers, then AM = ______.

10. The constant term in the expansion of (x + 1/x)⁶ is ______.

11. Σn² for n = 10 is ______.

12. If the first term of a GP is 3 and the common ratio is 1/3, then S∞ = ______.

III. True or False (with Corrections)

1. The number of terms in the expansion of (a + b)ⁿ is n.

False. The number of terms is n + 1 (r goes from 0 to n, giving n+1 terms).

2. In an AP, the sum of equidistant terms from the beginning and end is constant.

True. If a₁, a₂, …, aₙ is an AP, then a₁ + aₙ = a₂ + aₙ₋₁ = … = constant.

3. The infinite GP 1 + 2 + 4 + 8 + … converges to a finite sum.

4. If a, b, c are in HP, then 1/a, 1/b, 1/c are in AP.

True. By definition, the reciprocals of an HP form an AP.

5. AM ≥ GM ≥ HM holds for all real numbers.

False. The inequality AM ≥ GM ≥ HM holds only for positive real numbers. It does not hold when any number is negative or zero.

6. The binomial coefficient nCr equals nC(n−r).

True. This is the symmetry property of binomial coefficients: nCr = nC(n−r).

7. Σn³ = (Σn)² for the first n natural numbers.

True. Σn³ = [n(n+1)/2]² and Σn = n(n+1)/2, so Σn³ = (Σn)².

IV. Match the Following

• (A) → (2): Σn = n(n+1)/2
• (B) → (4): Σn² = n(n+1)(2n+1)/6
• (C) → (1): Σn³ = [n(n+1)/2]²
• (D) → (3): Σ nCr = 2ⁿ
• (E) → (5): General term T(r+1) = nCr · a^(n−r) · b^r
• (F) → (6): S∞ = a/(1 − r) for |r| < 1

V. Answer Briefly (2-Mark Questions)

Q1. Find the 7th term in the expansion of (2x + 3y)⁹.

Q2. Find the sum of the first 15 terms of the AP: 7, 13, 19, 25, …

Q3. Find the sum of the GP: 1 + 3 + 9 + … to 6 terms.

Q4. If the 3rd and 7th terms of an AP are 13 and 25 respectively, find the 15th term.

Q5. Prove that 2ⁿ > n for all n ∈ N.

By mathematical induction:

Hence, by induction, 2ⁿ > n for all n ∈ N.

VI. Answer in Detail (5-Mark Questions)

Q1. Expand (2x − 3y)⁵ using the Binomial Theorem and find the coefficient of x²y³.

The Binomial Theorem gives:

Here a = 2x, b = −3y, n = 5:

Coefficient of x²y³:

The coefficient of x²y³ is −1080.

Q2. The sum of the first three terms of a GP is 42 and their product is 512. Find the terms.

Let the three terms be a/r, a, ar (standard form for three GP terms).

Substituting a = 8 in (1):

Case 1: r = 4

Case 2: r = 1/4

The three terms are 2, 8, 32 (or equivalently 32, 8, 2).

Q3. Find the sum of the series 1² + 2² + 3² + … + 20² and prove that Σn² = n(n+1)(2n+1)/6 for n = 20.

Proof by induction:

Numerical calculation for n = 20:

Verification:

The sum is 2870.

Q4. Find the coefficient of the middle term in the expansion of (1 + x)¹⁰.

n = 10 is even, so there is one middle term: T(n/2 + 1) = T₆.

The coefficient of the middle term is 252.

Q5. Derive the formula Sₙ = n/2 [2a + (n−1)d] for an AP and use it to find the sum of the first 25 terms of the AP: 3, 7, 11, 15, …

Derivation:

Let the AP have first term a and common difference d. The terms are:

Writing the same sum in reverse:

Adding (1) and (2) term by term:

Application:

The sum of the first 25 terms is 1275.

8. Common Mistakes to Avoid

Mistake 1: Off-by-one in general term. T(r+1) = nCr · a^(n−r) · b^r. The index r starts from 0, not 1. The 3rd term means r = 2, not r = 3.

Mistake 2: Confusing AP sum formulas. Sₙ = n/2 [2a + (n−1)d] is different from Sₙ = n/2 (a + l). Use the first when you know a and d; use the second when you know the first and last terms.

Mistake 4: Wrong middle term count. If n is even, there is 1 middle term at position (n/2 + 1). If n is odd, there are 2 middle terms. Always check parity of n first.

Mistake 5: AM-GM-HM inequality conditions. The inequality AM ≥ GM ≥ HM applies only to positive numbers. If any number is zero or negative, the inequality may not hold.

Mistake 6: Confusing HP nth term. The nth term of an HP is NOT the nth term of the corresponding AP divided by n. It is the reciprocal of the nth term of the corresponding AP: aₙ(HP) = 1/(a + (n−1)d).

Mistake 7: Sign errors in binomial expansion with negative terms. When expanding (a − b)ⁿ, treat it as (a + (−b))ⁿ. Alternating signs come from (−b)^r = (−1)^r · b^r. Track signs carefully, especially for odd powers of r.

Mistake 8: Special series confusion. Σn = n(n+1)/2 and Σn³ = [n(n+1)/2]² are different. Note that Σn³ = (Σn)², but Σn² is NOT (Σn)². Σn² = n(n+1)(2n+1)/6.

9. Exam Checklist

Before walking into the exam, make sure you can do each of the following from memory:

• [ ] Write the Binomial Theorem expansion for (a + b)ⁿ and identify the general term
• [ ] Find a specific term in a binomial expansion using T(r+1) = nCr · a^(n−r) · b^r (remember: r starts at 0)
• [ ] Find the middle term(s) — check if n is even (1 middle term) or odd (2 middle terms)
• [ ] Find the constant term by setting the exponent of x to zero and solving for r
• [ ] Compute AP nth term: aₙ = a + (n − 1)d
• [ ] Compute AP sum: Sₙ = n/2 [2a + (n−1)d] or Sₙ = n/2 (a + l)
• [ ] Compute GP nth term: aₙ = ar^(n−1)
• [ ] Compute GP sum: Sₙ = a(rⁿ − 1)/(r − 1) and infinite sum S∞ = a/(1 − r) when |r| < 1
• [ ] Define HP and find the nth term by taking reciprocals of the corresponding AP
• [ ] State AM-GM-HM inequality: AM ≥ GM ≥ HM for positive numbers, and AM × HM = GM² for two numbers
• [ ] Recall special series: Σn = n(n+1)/2, Σn² = n(n+1)(2n+1)/6, Σn³ = [n(n+1)/2]²
• [ ] Verify answers: Plug back values into original formulas to catch arithmetic errors