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.
1
General term: T(r+1) = ⁸Cᵣ x⁸⁻ʳ 2ʳ. For T₅, r = 4.
2
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.
1
n = 6 (even). Middle term = T(n/2+1) = T(4).
✓ Middle term = 20a³b³
MediumFind the coefficient of x³ in (2x−1/x)⁹▾
Find the coefficient of x³ in (2x − 1/x)⁹.
1
General term: T(r+1) = ⁹Cᵣ (2x)⁹⁻ʳ (−1/x)ʳ = ⁹Cᵣ 2⁹⁻ʳ (−1)ʳ x⁹⁻ʳ⁻ʳ
2
Power of x = 9−2r. Set 9−2r = 3 → r = 3.
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.
1
T(r+1) = ¹²Cᵣ x¹²⁻ʳ (1/x²)ʳ = ¹²Cᵣ x¹²⁻ʳ⁻²ʳ = ¹²Cᵣ x¹²⁻³ʳ
2
For independence of x: 12−3r = 0 → r = 4.
✓ 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.
1
The trap: Students substitute x=0 (gives just the constant term) instead of x=1.
2
Sum of coefficients = value of the polynomial at x=1.
4
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~8 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.