The binomial theorem for NDA

The binomial theorem is one of the most predictable chapters in NDA maths. Almost every question asks for one specific piece of an expansion: a coefficient, a middle term, the term free of x, or the sum of all the coefficients. Writing out the whole expansion is slow and invites slips. The general term gets you straight to the piece you need, usually in under a minute.

Typical question types are:

• the number of terms in an expansion;
• the coefficient of a given power of x;
• the term independent of x (the constant term);
• the middle term or terms;
• sums of binomial coefficients, or of all the coefficients;
• remainders and divisibility using (1 + x)ⁿ, and simple approximations.

The theorem

For a positive integer n:

(a + b)ⁿ = nC0 aⁿ + nC1 aⁿ⁻¹b + nC2 aⁿ⁻²b² + … + nCn bⁿ

• There are n + 1 terms.
• In each term the powers of a and b add up to n.
• The coefficients read the same from both ends, because nCr = nC(n − r).

Why nCr? Expanding (a + b)ⁿ means choosing a or b from each of the n brackets. A term with bʳ arises every time b is chosen from r of the n brackets, and that can happen in nCr ways.

The general term

The (r + 1)th term is:

T(r+1) = nCr aⁿ⁻ʳ bʳ, for r = 0, 1, …, n.

Note the "r + 1": the 4th term has r = 3. For (a − b)ⁿ, the general term carries an extra factor (−1)ʳ.

The method for "find the term":

1. Write T(r+1) with the actual a and b, including signs and constants.
2. Collect all powers of x into a single power.
3. Set that power equal to the one you want (0 for the term independent of x) and solve for r.
4. If r is not a whole number between 0 and n, no such term exists.

Middle terms

The greatest binomial coefficient is the middle one: nC(n/2) for even n. For odd n, the two middle coefficients are equal and greatest.

Sums of coefficients

Put suitable values of x into (1 + x)ⁿ = nC0 + nC1 x + … + nCn xⁿ:

• x = 1: nC0 + nC1 + … + nCn = 2ⁿ.
• x = −1: nC0 − nC1 + nC2 − … = 0, so the sum of the even-position coefficients equals the sum of the odd-position ones, and each is 2ⁿ⁻¹.
• For any expansion in x, the sum of all the coefficients is the value at x = 1. For (3x − 2)⁵ that is (3 − 2)⁵ = 1.

Remainders and approximations

Remainders. Write the base as (a multiple of the divisor) ± 1, and expand. Every term except the last contains the divisor. For example, 2¹⁰⁰ = 2 × (2³)³³ = 2 × (7 + 1)³³, which is 2 × (a multiple of 7 + 1). The remainder on dividing by 7 is 2.

Approximations. For small x, (1 + x)ⁿ ≈ 1 + nx + nC2 x². For example, (1.02)⁶ ≈ 1 + 0.12 + 15 × 0.0004 = 1.126. The exact value is 1.1261… .

Worked NDA-style MCQs

Q1. The coefficient of x³ in (x² − 2/x)⁶ is:
(a) 160 (b) 240 (c) −240 (d) −160

T(r+1) = 6Cr (x²)⁶⁻ʳ (−2/x)ʳ = 6Cr (−2)ʳ x¹²⁻³ʳ. Set 12 − 3r = 3, so r = 3. The coefficient is 20 × (−8) = −160. Answer: (d).

Q2. The term independent of x in (x − 1/x²)⁹ is:
(a) 84 (b) 126 (c) −84 (d) −126

T(r+1) = 9Cr x⁹⁻ʳ (−1)ʳ x⁻²ʳ = 9Cr (−1)ʳ x⁹⁻³ʳ. Set 9 − 3r = 0, so r = 3. The term is 84 × (−1) = −84. Answer: (c).

Q3. The middle term of (3x − 1/x)⁴ is:
(a) 54 (b) −54 (c) 36 (d) 108

n = 4 gives 5 terms, so the middle one is T3 (r = 2). T3 = 4C2 (3x)² (−1/x)² = 6 × 9x² × (1/x²) = 54. Answer: (a).

Q4. The sum of the coefficients in the expansion of (3x − 2)⁵ is:
(a) 1 (b) 32 (c) 3,125 (d) −1

Put x = 1: (3 − 2)⁵ = 1. Answer: (a).

Q5. In (1 + x)ⁿ, the coefficients of x² and x³ are equal. Then n is:
(a) 4 (b) 5 (c) 6 (d) 7

nC2 = nC3, and since 2 ≠ 3, 2 + 3 = n. So n = 5. Check: 5C2 = 5C3 = 10. Answer: (b).

Q6. The remainder when 9⁵⁰ is divided by 8 is:
(a) 0 (b) 1 (c) 7 (d) 9

9⁵⁰ = (8 + 1)⁵⁰. Every term of the expansion except the last, 1, contains 8. Answer: (b).

Common mistakes

• Off-by-one in the term number. The 5th term is T(r+1) with r = 4.
• Dropping the sign. In (x − 2)ⁿ, (−2)ʳ changes sign with r.
• Forgetting the constants inside a bracket. In (2x + 3)ⁿ, the coefficient of a term includes powers of 2 and 3, not just nCr.
• Counting terms of a disguised expansion. (1 + 2x + x²)¹⁰ is (1 + x)²⁰, which has 21 terms, not 11.

Practice set

1. The number of terms in (1 + 2x + x²)¹⁰ is: (a) 11 (b) 20 (c) 21 (d) 30
2. The coefficient of x⁴ in (1 − x)⁷ is: (a) 35 (b) −35 (c) 21 (d) −21
3. The term independent of x in (2x + 1/x)⁶ is: (a) 20 (b) 60 (c) 160 (d) 240
4. The middle terms of (1 + x)⁷ are: (a) 35x⁴ only (b) 21x² and 35x³ (c) 70x⁴ only (d) 35x³ and 35x⁴
5. The sum of the coefficients in (1 − 2x)⁹ is: (a) 1 (b) −1 (c) 0 (d) 3⁹
6. 11C1 + 11C3 + 11C5 + … + 11C11 equals: (a) 2¹¹ (b) 2⁹ (c) 2¹¹ − 1 (d) 2¹⁰
7. The 3rd term of (x − 3)⁵ is: (a) −90x³ (b) 30x³ (c) 90x³ (d) 270x²
8. The greatest binomial coefficient in (1 + x)¹⁰ is: (a) 252 (b) 210 (c) 120 (d) 462

Answers:

1. (c) 21. The expression is (1 + x)²⁰.
2. (a) 35. 7C4 × (−1)⁴ = 35.
3. (c) 160. T(r+1) = 6Cr 2⁶⁻ʳ x⁶⁻²ʳ; r = 3 gives 20 × 8.
4. (d). n = 7 is odd, so T4 = 7C3 x³ and T5 = 7C4 x⁴, both with coefficient 35.
5. (b) −1. (1 − 2)⁹ = −1.
6. (d) 2¹⁰. The odd-position coefficients add to 2ⁿ⁻¹.
7. (c) 90x³. T3 = 5C2 x³ (−3)² = 10 × 9x³.
8. (a) 252. 10C5.

What to do next

• Practise the four-step general-term method until you can do it without writing the full expansion.
• Solve 25 questions from old NDA papers on this chapter, timed at 60 seconds each.
• Revise nCr in permutations and combinations; the binomial distribution in probability uses the same coefficients.