Binomial Theorem

Concept Core

Expansion, general term, middle term, and applications.

The Binomial Expansion

For any positive integer n:

(a+b)ⁿ = ⁿC₀aⁿ + ⁿC₁aⁿ⁻¹b + ⁿC₂aⁿ⁻²b² + ... + ⁿCₙbⁿ = Σᵣ₌₀ⁿ ⁿCᵣ aⁿ⁻ʳ bʳ

Total terms = n+1. Coefficients ⁿCᵣ are the binomial coefficients. Sum of all coefficients (put a=b=1): 2ⁿ.

General Term — T(r+1)

The (r+1)th term in the expansion of (a+b)ⁿ:

T(r+1) = ⁿCᵣ · aⁿ⁻ʳ · bʳ

r starts from 0. To find a specific term: set up T(r+1) and solve for r from the power condition. This formula solves 80% of binomial questions.

Middle Term(s)

When n is even: One middle term = T(n/2 + 1)

When n is odd: Two middle terms = T((n+1)/2) and T((n+3)/2)

Eg: (a+b)⁸ → middle term = T(5) = ⁸C₄ a⁴b⁴

Eg: (a+b)⁷ → middle terms = T(4) and T(5)

Coefficient of a Specific Power

To find coefficient of xᵏ in (ax+b)ⁿ:

Write general term: T(r+1) = ⁿCᵣ (ax)ⁿ⁻ʳ (b)ʳ

Power of x = n−r. Set n−r = k → r = n−k. Substitute r into T(r+1).

If the term is independent of x: set power of x = 0, solve for r.

Properties of Binomial Coefficients

Sum: ⁿC₀ + ⁿC₁ + ... + ⁿCₙ = 2ⁿ (put x=1 in (1+x)ⁿ)

Alternating sum: ⁿC₀ − ⁿC₁ + ⁿC₂ − ... = 0 (put x=−1)

Sum of even-index: = Sum of odd-index = 2ⁿ⁻¹

ⁿC₁ + 2·ⁿC₂ + ... = n·2ⁿ⁻¹ (differentiate (1+x)ⁿ)

Special Expansions (1+x)ⁿ for |x|<1

For fractional/negative n (infinite series):

(1+x)ⁿ = 1 + nx + n(n−1)/2! x² + n(n−1)(n−2)/3! x³ + ...

(1+x)⁻¹ = 1 − x + x² − x³ + ... (geometric series)
(1−x)⁻¹ = 1 + x + x² + x³ + ...

Formula Vault

Binomial theorem formulas for quick exam access.

Binomial Expansion

(a+b)ⁿ = Σ ⁿCᵣ aⁿ⁻ʳ bʳ

r from 0 to n; n+1 terms total

General Term

T(r+1) = ⁿCᵣ aⁿ⁻ʳ bʳ

r = 0,1,2,...,n

Middle Term (n even)

T(n/2 + 1)

Single middle term

Middle Terms (n odd)

T((n+1)/2) and T((n+3)/2)

Two middle terms

Sum of Coefficients

Σ ⁿCᵣ = 2ⁿ

Put a = b = 1

Alternating Sum

ⁿC₀ − ⁿC₁ + ... = 0

Put a=1, b=−1

Sum of Even/Odd Coefficients

Each = 2ⁿ⁻¹

Symmetric property

Binomial for |x|<1

(1+x)⁻¹ = 1−x+x²−...

Infinite geometric series

Worked Examples

5 problems — general term, coefficient finding, middle term, and trap.

EasyFind the 5th term in (x+2)⁸▾

Find T₅ in the expansion of (x+2)⁸.

General term: T(r+1) = ⁸Cᵣ x⁸⁻ʳ 2ʳ. For T₅, r = 4.

T₅ = ⁸C₄ · x⁴ · 2⁴ = 70 · x⁴ · 16 = 1120x⁴

✓ T₅ = 1120x⁴

EasyFind the middle term of (a+b)⁶▾

Find the middle term in the expansion of (a+b)⁶.

n = 6 (even). Middle term = T(n/2+1) = T(4).

T₄ = ⁶C₃ a³ b³ = 20a³b³

✓ Middle term = 20a³b³

MediumFind the coefficient of x³ in (2x−1/x)⁹▾

Find the coefficient of x³ in (2x − 1/x)⁹.

General term: T(r+1) = ⁹Cᵣ (2x)⁹⁻ʳ (−1/x)ʳ = ⁹Cᵣ 2⁹⁻ʳ (−1)ʳ x⁹⁻ʳ⁻ʳ

Power of x = 9−2r. Set 9−2r = 3 → r = 3.

T₄ = ⁹C₃ × 2⁶ × (−1)³ = 84 × 64 × (−1) = −5376

✓ Coefficient of x³ = −5376

EAPCET LevelFind the term independent of x in (x + 1/x²)¹²▾

Find the term independent of x (i.e., x⁰) in (x + 1/x²)¹².

T(r+1) = ¹²Cᵣ x¹²⁻ʳ (1/x²)ʳ = ¹²Cᵣ x¹²⁻ʳ⁻²ʳ = ¹²Cᵣ x¹²⁻³ʳ

For independence of x: 12−3r = 0 → r = 4.

T₅ = ¹²C₄ = 495

✓ Term independent of x = 495

Trap QuestionSum of coefficients of (3x−2)⁸ — students use wrong substitution▾

Find the sum of all coefficients in (3x−2)⁸. ⚠️ Common substitution error.

The trap: Students substitute x=0 (gives just the constant term) instead of x=1.

Sum of coefficients = value of the polynomial at x=1.

(3(1)−2)⁸ = (1)⁸ = 1

Verify: Sum of binomial coefficients (ⁿCᵣ) = 2ⁿ = 256, but these get multiplied by powers of 3 and −2. Putting x=1 accounts for all coefficient multipliers.

✓ Sum of coefficients = 1

Mistake DNA

3 critical errors in Binomial Theorem questions.

🔢

Using Wrong r for T(r+1)

Students use r when the question asks for the r-th term, not (r+1)-th. Off-by-one error everywhere.

❌ Wrong

4th term: use r=4 T₄ = ⁿC₄... ✗ (T(r+1) means r=3 for 4th term)

✓ Correct

T(r+1)=T₄ → r=3 T₄ = ⁿC₃ aⁿ⁻³ b³ ✓ Always: Tₖ uses r=k−1

The formula T(r+1) means: when r=0 you get T1, when r=1 you get T2, etc. For the kth term, r = k−1.

📐

Wrong Middle Term for Odd n

When n is odd, there are TWO middle terms. Students pick only one.

❌ Wrong

(a+b)⁷ has one middle term T₄ ✗ (n=7 is odd → two middle terms)

✓ Correct

n=7 odd: T((7+1)/2)=T₄ AND T((7+3)/2)=T₅ ✓ Both are middle terms

n odd → two middle terms at positions (n+1)/2 and (n+3)/2. n even → one middle term at position n/2+1.

🎯

Sum of Coefficients: Substituting x=0 Instead of x=1

x=0 gives only the constant term, not the sum of all coefficients.

❌ Wrong

Sum of coeff of (2x+3)⁵: put x=0: 3⁵=243 ✗ (just the constant term)

✓ Correct

Put x=1: (2+3)⁵=5⁵ =3125 ✓ All coefficients summed

Sum of coefficients means: what is the total when all x terms are just 1? Substitute x=1 into the expression.

Chapter Intelligence

Binomial theorem is directly linked to Permutations, and feeds into series problems.

EAPCET Weightage (2019–2024)

General term / specific term

Term independent of x

Middle term

Coefficient of xⁿ

Sum of coefficients

High-Yield PYQ Patterns

Find T(r+1) for specific rTerm independent of xCoefficient of x³ in (ax+b/x)ⁿMiddle term when n is even/oddSum of all binomial coefficientsNumerically greatest term

Exam Strategy

Start every binomial question by writing T(r+1) = ⁿCᵣ aⁿ⁻ʳ bʳ. Then substitute what's given and identify r from the power condition.
Term independent of x: set power of x = 0, solve for r. If r is not a whole number, that term doesn't exist.
Middle term: check if n is even or odd first. This determines how many middle terms exist.
Sum of coefficients of (f(x))ⁿ: always substitute x = 1 into f(x), then raise to n. One step, no expansion needed.
Binomial coefficients are the same as combinations: ⁿCᵣ. If you know P&C well, this chapter's formulas are already familiar.

Concept Core

Formula Vault

Worked Examples

Mistake DNA

Chapter Intelligence

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