Binomial Theorem — JEE Main Mathematics Notes & Formulas

1. Chapter at a Glance

The Binomial Theorem tells us how to expand (a + b)ⁿ without multiplying the bracket n times. It connects directly with combinations: the coefficients are the numbers ⁿC_r. JEE Main usually asks one or two questions — on a particular term or coefficient, the term independent of x, sums of binomial coefficients and remainders.

What JEE asks most

The general term, the middle term and the term independent of x.
Coefficient of a power of x in a product such as (1 + x + x²)(1 − x)ⁿ.
Sums like ΣⁿC_r, Σr·ⁿC_r and Σ(ⁿC_r)².
Remainder or last digits of large powers such as 7¹⁰³ or 3¹⁰⁰.

2. The Binomial Theorem

For a positive integer n,

(a + b)ⁿ = ⁿC₀aⁿ + ⁿC₁aⁿ⁻¹b + ⁿC₂aⁿ⁻²b² + … + ⁿC_nbⁿ

There are n + 1 terms. In each term, the powers of a and b add up to n.
Coefficients equidistant from the two ends are equal, because ⁿC_r = ⁿC_n−r.
For (a − b)ⁿ, the terms alternate in sign.

Pascal's triangle: row n gives the coefficients of (a + b)ⁿ. The rule ⁿC_r + ⁿC_r−1 = ⁿ⁺¹C_r builds each row.

3. General Term, Middle Term and Independent Term

T_r+1 = ⁿC_r a^n−r b^r, r = 0, 1, …, n

Note that T_r+1 is the (r + 1)-th term — the index is one more than r.

Middle term

If n is even, there is one middle term: T_{n/2 + 1}.
If n is odd, there are two middle terms: T_(n+1)/2 and T_(n+3)/2.

Term independent of x

Write the general term, collect the power of x, and set it equal to 0. For (ax^p + b/x^q)ⁿ, the power of x in T_r+1 is p(n − r) − qr; solve p(n − r) − qr = 0 for r. If r is not a whole number from 0 to n, there is no independent term.

Greatest term and greatest coefficient

The greatest binomial coefficient is the middle one: ⁿC_n/2 (n even) or ⁿC_(n±1)/2 (n odd).
For the numerically greatest term of (1 + x)ⁿ, compare T_r+1/T_r = (n − r + 1)|x|/r with 1.

4. Properties of Binomial Coefficients

Write C_r for ⁿC_r. Putting special values of x in (1 + x)ⁿ = C₀ + C₁x + … + C_nxⁿ gives many results:

Sum	Value	How
C₀ + C₁ + … + C_n	2ⁿ	x = 1
C₀ − C₁ + C₂ − …	0	x = −1
C₀ + C₂ + C₄ + …	2ⁿ⁻¹	add the two above
C₁ + C₃ + C₅ + …	2ⁿ⁻¹	subtract
Σ r C_r	n 2ⁿ⁻¹	differentiate, x = 1
Σ (r + 1) C_r	(n + 2) 2ⁿ⁻¹	add 2ⁿ to the above
Σ C_r/(r + 1)	(2ⁿ⁺¹ − 1)/(n + 1)	integrate from 0 to 1
Σ C_r²	²ⁿC_n	coefficient of xⁿ in (1 + x)²ⁿ

Useful identity

r · ⁿC_r = n · ⁿ⁻¹C_r−1. Use it to "absorb" the r in sums like Σ r·ⁿC_r.

5. Remainders, Last Digits and the Multinomial Case

To find the remainder of a large power, rewrite the base close to a multiple of the divisor. For example 7² = 49 = 50 − 1, so 7² ≡ −1 (mod 25). Then use the binomial expansion or simply powers of −1.

Last two digits of a number = remainder on division by 100.
(1 + x)ⁿ ≥ 1 + nx for x > −1 (Bernoulli's inequality) — useful for comparing large numbers.

Multinomial expansion

The number of terms in (x₁ + x₂ + … + x_r)ⁿ is ^n+r−1C_r−1. So (a + b + c)¹⁰ has ¹²C₂ = 66 terms.

Binomial series for any index (for reference)

For |x| < 1 and any real n: (1 + x)ⁿ = 1 + nx + n(n − 1)x²/2! + … In particular (1 − x)⁻¹ = 1 + x + x² + … and (1 − x)⁻² = 1 + 2x + 3x² + …

6. Formula Master Table

Result	Formula
General term	T_r+1 = ⁿC_r aⁿ⁻ʳ bʳ
Number of terms	n + 1
Middle term (n even)	T_n/2+1
r-th term from the end	(n − r + 2)-th term from the beginning
Sum of coefficients	Put every variable = 1
Absorption identity	r·ⁿC_r = n·ⁿ⁻¹C_r−1
Sum of squares	Σ(ⁿC_r)² = ²ⁿC_n
Vandermonde's identity	Σ ᵐC_k ⁿC_r−k = ^m+nC_r
Terms in multinomial	^n+r−1C_r−1
Ratio of successive terms	T_r+1/T_r = (n − r + 1)b/(ra)

7. Mind Map

Binomial Theorem

Expansion

n + 1 terms
Pascal's triangle
Symmetric coefficients

Special terms

General T_r+1
Middle term(s)
Independent of x

Coefficient sums

x = 1, x = −1
Differentiate / integrate
Σ C_r² = ²ⁿCₙ

Applications

Remainders
Last digits
Multinomial terms

8. Tips and Tricks

Shortcuts

For the coefficient of x^k in (1 + x + x²)(1 − x)ⁿ, add the coefficients of x^k, x^k−1 and x^k−2 in (1 − x)ⁿ.
If a sum has r in front of ⁿC_r, think "differentiate"; if it has 1/(r + 1), think "integrate".
For remainders, look for a power of the base that is ±1 modulo the divisor (e.g. 2⁵ = 32 ≡ −1 mod 33, 3⁴ = 81 ≡ 1 mod 80).
The sum of coefficients of any polynomial in x is its value at x = 1 — no expansion needed.

Mistakes students make

Confusing the r-th term with T_r = ⁿC_r…; the r-th term is T_r = ⁿC_r−1….
Forgetting the sign in (a − b)ⁿ: the term contains (−b)ʳ.
Calling ⁿC_r "the coefficient of the term" when the term also contains numbers from a and b (e.g. in (2x − 3)ⁿ).

9. Important One-liners

(a + b)ⁿ + (a − b)ⁿ contains only the odd-numbered terms, doubled.
The coefficient of xʳ in (1 + x)ⁿ is ⁿC_r; in (1 − x)ⁿ it is (−1)ʳ ⁿC_r.
The binomial coefficients in the expansion first increase and then decrease.
2ⁿ is the total number of subsets and also the sum of all binomial coefficients.
11ⁿ for small n gives the rows of Pascal's triangle (11² = 121, 11³ = 1331, 11⁴ = 14641).
If the coefficients of three consecutive terms are in AP, a quadratic in n or r results.
The number of terms in (x + a)ⁿ + (x − a)ⁿ is (n/2 + 1) if n is even and (n + 1)/2 if n is odd.

10. PYQ-pattern Questions with Solutions

These questions are original and modelled on concepts frequently tested in JEE Main.

Q1Numerical

Find the term independent of x in the expansion of (x² + 1/x)⁹.

Show solution

T_r+1 = ⁹C_r (x²)^9−r(1/x)ʳ = ⁹C_r x^18−3r.

For the independent term, 18 − 3r = 0 ⇒ r = 6. The term is ⁹C₆ = 84.

Answer: 84

Q2Medium

The coefficient of x⁴ in (1 + x + x²)(1 − x)⁶ is:

15
10
−20
5

Show solution

The coefficients of (1 − x)⁶ are 1, −6, 15, −20, 15, −6, 1.

Coefficient of x⁴ = (coeff of x⁴) + (coeff of x³) + (coeff of x²) = 15 + (−20) + 15 = 10.

Answer: (B) 10

Q3Medium

The middle term in the expansion of (2x − 1/(2x))¹⁰ is:

252
−252
−126
126

Show solution

n = 10 is even, so the middle term is T₆ (r = 5).

T₆ = ¹⁰C₅(2x)⁵(−1/(2x))⁵ = 252 × 32x⁵ × (−1/(32x⁵)) = −252.

Answer: (B)

Q4Numerical

Find the value of ¹⁰C₀ + 2·¹⁰C₁ + 3·¹⁰C₂ + … + 11·¹⁰C₁₀.

Show solution

Σ(r + 1)ⁿC_r = Σ r·ⁿC_r + ΣⁿC_r = n·2ⁿ⁻¹ + 2ⁿ = (n + 2)2ⁿ⁻¹.

For n = 10: 12 × 2⁹ = 12 × 512 = 6144.

Answer: 6144

Q5Numerical

Find the remainder when 7¹⁰³ is divided by 25.

Show solution

7² = 49 = 50 − 1 ≡ −1 (mod 25), so 7⁴ ≡ 1 (mod 25).

103 = 4 × 25 + 3, so 7¹⁰³ ≡ 7³ = 343 (mod 25). 343 = 13 × 25 + 18.

Answer: 18

Q6Easy

The number of terms in the expansion of (a + b + c)¹⁰ is:

Show solution

For (x₁ + x₂ + x₃)ⁿ the number of terms is ⁿ⁺²C₂ = ¹²C₂ = 66.

Answer: (C)

Q7Hard

If the coefficients of the 5th, 6th and 7th terms of (1 + x)ⁿ are in AP, then n can be:

7 or 14
6 or 12
5 or 10
8 or 16

Show solution

The coefficients are ⁿC₄, ⁿC₅, ⁿC₆. AP condition: 2·ⁿC₅ = ⁿC₄ + ⁿC₆.

Divide by ⁿC₅ and use ⁿC₄/ⁿC₅ = 5/(n − 4) and ⁿC₆/ⁿC₅ = (n − 5)/6:

2 = 5/(n − 4) + (n − 5)/6 ⇒ 12(n − 4) = 30 + (n − 5)(n − 4) ⇒ n² − 21n + 98 = 0 ⇒ n = 7 or 14.

Answer: (A)

Q8Hard

The last two digits of 3¹⁰⁰ are:

Show solution

3¹⁰⁰ = (3²)⁵⁰ = 9⁵⁰ = (10 − 1)⁵⁰. Expanding, every term except the last two contains 10² = 100:

(10 − 1)⁵⁰ ≡ ⁵⁰C₄₉·10·(−1)⁴⁹ + (−1)⁵⁰ = −500 + 1 ≡ 1 (mod 100).

So the last two digits are 01.

Answer: (A)

11. Quick Revision Summary

Remember before the exam

T_r+1 = ⁿC_raⁿ⁻ʳbʳ; there are n + 1 terms.
Independent term: set the power of x in T_r+1 equal to 0.
ΣC_r = 2ⁿ, Σ even = Σ odd = 2ⁿ⁻¹, Σ rC_r = n2ⁿ⁻¹, ΣC_r² = ²ⁿC_n.
Remainders: make the base ±1 modulo the divisor.
Terms in (a + b + c)ⁿ: ⁿ⁺²C₂.

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