Binomial Theorem

The binomial theorem is a compact, high-yield NDA topic — 2–3 questions, mostly on the general term and coefficients. Once you can write the (r+1)th term, most items become one substitution.

Core ideas

Expansion: (a + b)ⁿ = Σ (from r=0 to n) ⁿCᵣ a^(n−r) bʳ. There are n+1 terms.

General term: T_(r+1) = ⁿCᵣ a^(n−r) bʳ. Almost every question is 'find a particular term / coefficient / term independent of x' — set the power of x you need and solve for r.

Middle term: if n is even, one middle term, the (n/2 + 1)th; if n is odd, two middle terms, the (n+1)/2 th and (n+3)/2 th.

Term independent of x (constant term): set the exponent of x in T_(r+1) equal to 0 and solve for r.

Key sums: ⁿC₀ + ⁿC₁ + … + ⁿCₙ = 2ⁿ (put a=b=1). Sum of even-position = sum of odd-position = 2^(n−1).

Greatest coefficient is the middle binomial coefficient ⁿC_(n/2).

Exam Tricks & Tips

• 🎯 Always write T_(r+1) first; the (r+1) indexing (not r) prevents off-by-one errors in 'which term' questions.
• 🎯 For 'coefficient of xᵏ', form the exponent of x in T_(r+1), set = k, solve for r, then plug back.
• 🎯 Put a = b = 1 to get ΣⁿCᵣ = 2ⁿ; put a = 1, b = −1 to get the alternating sum = 0.
• 🎯 (1+x)ⁿ has T_(r+1) = ⁿCᵣ xʳ — the coefficient of xʳ is simply ⁿCᵣ.
• 🎯 For number-of-terms in (a+b+c)ⁿ use ⁿ⁺²C₂ (multinomial), not n+1.
• ❌ Common mistake: taking the middle term as the (n/2)th instead of (n/2 + 1)th for even n — always add one.

Expected exam pattern

Typical items: find the coefficient of a given power, the term independent of x, the middle term, or evaluate a coefficient sum. Occasionally a divisibility/remainder question using (1+x)ⁿ.

Quick recap

Master T_(r+1) = ⁿCᵣ a^(n−r) bʳ, the middle-term rule, the constant-term method (exponent = 0), and ΣⁿCᵣ = 2ⁿ. Index terms as (r+1) to avoid off-by-one slips.

Cover, recall, check. 10 cards on binomial expansion.

Q1. Binomial expansion of (a+b)ⁿ?
A1. Σ ⁿCᵣ a^(n−r) bʳ, r = 0 to n; it has n+1 terms.

Q2. General (r+1)th term?
A2. T_(r+1) = ⁿCᵣ a^(n−r) bʳ.

Q3. Middle term when n is even?
A3. Single middle term: the (n/2 + 1)th term.

Q4. Middle terms when n is odd?
A4. Two: the (n+1)/2 th and (n+3)/2 th terms.

Q5. How to find the term independent of x?
A5. Set the exponent of x in T_(r+1) to 0, solve for r.

Q6. Coefficient of xʳ in (1+x)ⁿ?
A6. ⁿCᵣ.

Q7. Sum ⁿC₀ + ⁿC₁ + … + ⁿCₙ?
A7. 2ⁿ (put a = b = 1).

Q8. Sum of even-indexed = sum of odd-indexed coefficients = ?
A8. 2^(n−1) each.

Q9. Greatest binomial coefficient in (1+x)ⁿ?
A9. The middle one, ⁿC_(n/2) for even n.

Q10. Number of terms in (a+b+c)ⁿ?
A10. ⁿ⁺²C₂.

Raising (a + b) to a large power by hand is hopeless — the Binomial Theorem gives every term in one formula, and NDA loves testing individual terms.

Core idea / what this tests: The expansion of (a + b)ⁿ, extracting a specific or general term, and reading off coefficients — including the middle and greatest term.

Deep explanation

Beginner — The expansion

For a positive integer n:
(a + b)ⁿ = Σ ⁿCᵣ · a^(n−r) · bʳ, r = 0 to n.
There are n + 1 terms, coefficients are the binomial coefficients ⁿC₀, ⁿC₁, …, ⁿCₙ (Pascal's triangle), and powers of a decrease while powers of b increase, their sum always n.

Intermediate — The general term

The (r+1)-th term is Tᵣ₊₁ = ⁿCᵣ · a^(n−r) · bʳ. This single expression answers most questions: to find the term with a chosen power of x, set the exponent equal to that power and solve for r.
Sum of all coefficients = 2ⁿ (put a = b = 1). Sum of odd-position = sum of even-position = 2^(n−1).

Advanced — Middle and independent terms

• Middle term: if n is even, one middle term T_(n/2 + 1); if n is odd, two middle terms.
• Term independent of x ("constant term"): set the net power of x to zero in the general term and solve for r.
• Greatest coefficient sits at the middle. For the greatest term numerically, use the ratio T_(r+1)/T_r.

Worked example

Find the coefficient of x⁵ in (x + 2)^8.
General term: T_(r+1) = ⁸Cᵣ · x^(8−r) · 2ʳ. For x⁵ we need 8 − r = 5 ⇒ r = 3.
Coefficient = ⁸C₃ · 2³ = 56 · 8 = 448.

NDA relevance

Expect a "find the term / coefficient of xᵏ", a "term independent of x", and occasionally a "sum of coefficients" or "middle term" question. All are one application of the general-term formula — very high reward per second.

Exam tricks & shortcuts

• Write the general term first, before anything else — it converts every question into "solve for r".
• Sum of coefficients: substitute the variable = 1.
• For (1 + x)ⁿ, the coefficient of xʳ is simply ⁿCᵣ.
• ⁿCᵣ symmetry (ⁿCᵣ = ⁿC₍ₙ₋ᵣ₎) can simplify awkward r values.

Key formulas

• (a+b)ⁿ = Σ_{r=0}^{n} ⁿCr a^{n−r} b^r; number of terms n+1.
• General term T_{r+1} = ⁿCr a^{n−r} b^r.
• Middle term: (n/2+1)-th if n even; two middle terms if n odd.
• Term independent of x: make the power of x zero in T_{r+1}.
• (1+x)ⁿ = Σ ⁿCr xʳ; Σ ⁿCr = 2ⁿ.
• Greatest coefficient is the middle term.

Usage: write the general term, isolate the power of x, and solve for r.

Worked Example

Problem: Find the term independent of x in the expansion of (x² − 1/x)⁹.

Solution:
The general (r+1)th term of (a + b)ⁿ is T_(r+1) = C(n, r) · a^(n−r) · b^r.
Here a = x², b = −1/x, and n = 9:
T_(r+1) = C(9, r) · (x²)^(9−r) · (−1/x)^r
= C(9, r) · (−1)^r · x^(2(9−r)) · x^(−r)
= C(9, r) · (−1)^r · x^(18 − 2r − r)
= C(9, r) · (−1)^r · x^(18 − 3r).

The term independent of x has exponent 0:
18 − 3r = 0 ⇒ r = 6.

So the required term is T₇ = C(9, 6) · (−1)⁶ · x⁰ = C(9, 6).
C(9, 6) = C(9, 3) = (9 × 8 × 7)/(3 × 2 × 1) = 504/6 = 84.

Answer: The term independent of x is 84.