Binomial Theorem: JEE Main Maths formulas and notes

Q: How do I solve “Coefficient or term independent of x” questions in JEE Main?

A binomial with powers of x is given. Find the coefficient of a power of x, or the term without x. 1. Write the general term T_(r+1) = ⁿC_r xⁿ⁻ʳ aʳ with the actual expressions. 2. Add the powers of x in the term and equate to the required power. 3. Solve for r, check that r is a whole number from 0 to n, and put it back. Example: (x² − 1/x)⁹ has its constant term at r = 6, equal to 84.

Q: How do I solve “Sum of binomial coefficients” questions in JEE Main?

A sum with ⁿC_r, possibly with r, signs or squares is given. 1. Compare it with (1 + x)ⁿ and choose x = 1 or x = −1. 2. If a factor r appears, differentiate (1 + x)ⁿ first and then put x = 1. 3. Use Σ(ⁿC_r)² = ²ⁿCₙ for squares. Example: Σ r·¹⁰C_r = 10 × 2⁹ = 5120.

Q: How do I solve “Remainder when a power is divided” questions in JEE Main?

A large power is divided by a small number. Find the remainder or show divisibility. 1. Write the base as a multiple of the divisor plus or minus a small number. 2. Expand with the binomial theorem; all but the last one or two terms are multiples of the divisor. 3. Read the remainder from the leftover terms and add the divisor if it is negative. Example: 7¹⁰⁰ leaves remainder 1 when divided by 6.

Q: How do I solve “Middle term, greatest term and coefficient relations” questions in JEE Main?

Find the middle term or the greatest term, or use a ratio of consecutive coefficients to find n. 1. Use T_(n/2 + 1) for even n, two middle terms for odd n. 2. For the greatest term compute m = (n + 1)|x|/(1 + |x|) and take its integer part. 3. For ratio problems use ⁿC_r/ⁿC_(r−1) = (n − r + 1)/r. Example: (1 + 2)¹⁰: the greatest term is T₈.

Binomial Theorem gives 1 to 2 questions in JEE Mains in most years: a coefficient or a term independent of x, a sum of binomial coefficients, or a remainder. About 6 hours in 4 days is enough. The idea of choosing, ⁿC_r, is also the base of probability, so the effort pays again.

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Quick revision

(x + a)ⁿ has n + 1 terms; T_(r+1) = ⁿC_r xⁿ⁻ʳ aʳ
Σ ⁿC_r = 2ⁿ; alternating sum 0; even-place sum = odd-place sum = 2ⁿ⁻¹
Σ r·ⁿC_r = n·2ⁿ⁻¹; Σ(ⁿC_r)² = ²ⁿCₙ
middle term: one if n even, two if n odd
remainder: write the base as (multiple of d) ± 1 and expand

Formulas with solved examples

Expansion and general term

Binomial theorem for a positive integer n

(x + a)ⁿ = ⁿC₀xⁿ + ⁿC₁xⁿ⁻¹a + ⁿC₂xⁿ⁻²a² + … + ⁿCₙaⁿ
n + 1 terms
the powers of x and a add up to n in every term
(1 + x)ⁿ = Σ ⁿC_r xʳ

How to apply. Write each term as a coefficient, a power of the first quantity and a power of the second. Replace a by −a to get alternating signs.

Example. (1 + x)¹⁰ has 11 terms and the coefficient of x³ is ¹⁰C₃ = 120.

General term

T_(r+1) = ⁿC_r xⁿ⁻ʳ aʳ
for (x + a)ⁿ the (r + 1)th term from the end is T_(n−r+1)
the rth term is T_r = ⁿC_(r−1) xⁿ⁻ʳ⁺¹ aʳ⁻¹

How to apply. To find a particular term, put the required power of x equal to the power in T_(r+1) and solve for r. Check that r is a whole number from 0 to n.

Example. In (x² − 1/x)⁹ the general term is ⁹C_r (−1)ʳ x^18 − 3r. The power 0 needs r = 6, so the term is ⁹C₆ = 84.

Term independent of x

set the power of x in T_(r+1) equal to 0 and solve for r
if r is not a whole number between 0 and n there is no such term

How to apply. Collect the powers of x from both parts of the term, add them, and set the total to zero.

Example. (x + 1/x)⁶: the power is 6 − 2r, which is 0 for r = 3. The term is ⁶C₃ = 20.

Rational terms

in (aᵖ + bᑫ)ⁿ with fractional powers, T_(r+1) is rational when every fractional power becomes an integer
list the values of r from 0 to n that work and count or add them

How to apply. Write the term with exponents r/p and (n − r)/q. Find the r that make both exponents whole numbers.

Middle and greatest term

Middle term

n even: one middle term, T_(n/2 + 1)
n odd: two middle terms, T_((n+1)/2) and T_((n+3)/2)
the middle coefficient of (1 + x)ⁿ is the largest coefficient

How to apply. Decide whether n is even or odd first, then write the term number and use the general term.

Example. (1 + x)¹⁰ has the single middle term T₆ with coefficient ¹⁰C₅ = 252.

Greatest term of (1 + x)ⁿ

T_(r+1)/T_r = ((n − r + 1)/r)
x
the terms rise while this ratio is above 1
let m = (n + 1)|x|/(1 + |x|)
if m is not a whole number the greatest term is T_(⌊m⌋+1)
if m is a whole number then T_m and T_(m+1) are equal and greatest

How to apply. Find m, take its integer part, and use it to name the term. Use absolute values when x is negative.

Example. (1 + 2)¹⁰ terms: m = 11 × 2/3 = 7.33, so the greatest term is T₈ (checked: the 8th term is the largest).

Coefficients and properties

Coefficient of a given power of x

write T_(r+1)
equate the power of x to the required power
solve for r
the coefficient is ⁿC_r times the numbers that come with it
for a product (1 + x)ᵐ(1 + x)ⁿ add the products of matching coefficients

How to apply. For a product of two expansions, list the pairs of powers that add up to the target and add their coefficient products.

Example. In (1 + x)ᵐ(1 + x)ⁿ = (1 + x)ᵐ⁺ⁿ the coefficient of xʳ is ᵐ⁺ⁿC_r.

Properties of ⁿC_r

ⁿC_r = ⁿC_(n−r)
ⁿC_x = ⁿC_y ⇒ x = y or x + y = n
ⁿC_r + ⁿC_(r−1) = ⁿ⁺¹C_r
ⁿC_r = (n/r) ⁿ⁻¹C_(r−1)
ⁿC_(r)/ⁿC_(r−1) = (n − r + 1)/r

How to apply. Use the ratio form when three consecutive coefficients are in a given ratio. Use the symmetry to match equal coefficients.

Example. If ¹⁰C_a = ¹⁰C_b with a ≠ b then a + b = 10.

Sums of binomial coefficients

Standard sums

ⁿC₀ + ⁿC₁ + … + ⁿCₙ = 2ⁿ
ⁿC₀ − ⁿC₁ + ⁿC₂ − … = 0
sum of even-place coefficients = sum of odd-place coefficients = 2ⁿ⁻¹
ΣrⁿC_r = n·2ⁿ⁻¹
Σ(ⁿC_r)² = ²ⁿCₙ

How to apply. Put x = 1 or x = −1 in (1 + x)ⁿ. To bring in a factor r, differentiate (1 + x)ⁿ and then put x = 1. To get 1/(r + 1), integrate.

Example. Σ r·¹⁰C_r = 10 × 2⁹ = 5120 and Σ(⁸C_r)² = ¹⁶C₈ = 12870.

Vandermonde and related identities

ᵐCᵣ·ⁿC₀ + ᵐCᵣ₋₁·ⁿC₁ + … + ᵐC₀·ⁿCᵣ = ᵐ⁺ⁿCᵣ
ⁿC₀·ⁿCᵣ + ⁿC₁·ⁿCᵣ₊₁ + … = ²ⁿCₙ₊ᵣ
Σ ⁿC_r/(r + 1) = (2ⁿ⁺¹ − 1)/(n + 1)

How to apply. Compare coefficients of the same power on both sides of (1 + x)ᵐ(1 + x)ⁿ = (1 + x)ᵐ⁺ⁿ.

Remainders, divisibility and approximation

Remainders and divisibility

write the number as (a ± 1)ⁿ or (multiple of d + k)ⁿ
expand
every term except the last two or three is a multiple of d
read the remainder from what is left
(1 + k)ⁿ = 1 + nk + (multiple of k²)

How to apply. Break the base into a nearby multiple of the divisor plus a small number, then expand and keep only the terms that are not multiples.

Example. 7¹⁰⁰ ÷ 6: 7 = 6 + 1, so 7¹⁰⁰ = 1 + (multiple of 6). Remainder 1. And 11²⁵ ÷ 9 = (9 + 2)²⁵, which is 2²⁵ mod 9. Remainder 2.

Approximation and last digits

(1 + x)ⁿ ≈ 1 + nx when |x| is very small
for the last digits work with (10 − 1)ⁿ or (100 ± 1)ⁿ and keep only the last terms

How to apply. Keep the first two or three terms when the later ones are too small for the digits asked.

Example. (1.01)¹⁰ ≈ 1 + 0.10 + 0.0045 = 1.1045, close to the exact 1.1046.

Any index and multinomial

Expansion when n is not a positive integer

(1 + x)ⁿ = 1 + nx + n(n − 1)/2! x² + n(n − 1)(n − 2)/3! x³ + … valid for |x| < 1
infinite series
(1 − x)⁻¹ = 1 + x + x² + …
(1 − x)⁻² = 1 + 2x + 3x² + …

How to apply. Use it when n is negative or a fraction. The number of terms never ends, and x must be small in size.

Example. (1 − x)⁻² = 1 + 2x + 3x² + …, so the coefficient of x⁹ is 10.

Multinomial theorem

in (x₁ + x₂ + … + x_k)ⁿ the coefficient of x₁^a₁ … x_k^a_k is n!/(a₁! … a_k!) with a₁ + … + a_k = n
the number of terms is ⁿ⁺ᵏ⁻¹C_(k−1)

How to apply. Use it when there are three or more terms inside the bracket. Count the solutions of a₁ + … + a_k = n for the number of terms.

Example. (1 + x + x²)ⁿ has 2n + 1 terms.

Most asked question types

How do I solve “Coefficient or term independent of x” questions in JEE Main?

A binomial with powers of x is given. Find the coefficient of a power of x, or the term without x. 1. Write the general term T_(r+1) = ⁿC_r xⁿ⁻ʳ aʳ with the actual expressions. 2. Add the powers of x in the term and equate to the required power. 3. Solve for r, check that r is a whole number from 0 to n, and put it back. Example: (x² − 1/x)⁹ has its constant term at r = 6, equal to 84.

How do I solve “Sum of binomial coefficients” questions in JEE Main?

A sum with ⁿC_r, possibly with r, signs or squares is given. 1. Compare it with (1 + x)ⁿ and choose x = 1 or x = −1. 2. If a factor r appears, differentiate (1 + x)ⁿ first and then put x = 1. 3. Use Σ(ⁿC_r)² = ²ⁿCₙ for squares. Example: Σ r·¹⁰C_r = 10 × 2⁹ = 5120.

How do I solve “Remainder when a power is divided” questions in JEE Main?

A large power is divided by a small number. Find the remainder or show divisibility. 1. Write the base as a multiple of the divisor plus or minus a small number. 2. Expand with the binomial theorem; all but the last one or two terms are multiples of the divisor. 3. Read the remainder from the leftover terms and add the divisor if it is negative. Example: 7¹⁰⁰ leaves remainder 1 when divided by 6.

How do I solve “Middle term, greatest term and coefficient relations” questions in JEE Main?

Find the middle term or the greatest term, or use a ratio of consecutive coefficients to find n. 1. Use T_(n/2 + 1) for even n, two middle terms for odd n. 2. For the greatest term compute m = (n + 1)|x|/(1 + |x|) and take its integer part. 3. For ratio problems use ⁿC_r/ⁿC_(r−1) = (n − r + 1)/r. Example: (1 + 2)¹⁰: the greatest term is T₈.

Common myths and the truth

The (r + 1)th term has ⁿC_(r+1). It has ⁿC_r. The term number is one more than r.
All binomial expansions are finite. Only for a positive integer n. A negative or fractional index gives an infinite series for |x| < 1.

Mistakes to avoid

Using T_r in place of T_(r+1).
Forgetting the sign when a is negative.
Missing that r must be a whole number from 0 to n.
Taking n even or odd wrongly for the middle term.

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