Mathematical Induction and Binomial Theorem – JEE Main Maths Formula Sheet & Class 11 Notes | All Formulas & Results

Principle of Mathematical Induction (PMI) — Framework and Standard Results

Why the PMI Framework and Its Standard Results Are JEE Main Maths Foundations

The Principle of Mathematical Induction (PMI) states: to prove that a proposition P(n) is true for all positive integers n ≥ n₀, it is sufficient to prove: (i) P(n₀) is true (Base case), and (ii) whenever P(k) is true for some integer k ≥ n₀, then P(k+1) is also true (Inductive step). Both steps together guarantee P(n) is true for all n ≥ n₀ by the Well-Ordering Principle of natural numbers.

Three steps of PMI (Mathematical Induction framework for JEE Main):

Step 1 — Base Case: Verify P(1) (or P(n₀)) is true. Substitute n=1 (or n=n₀) into the statement and confirm.

Step 2 — Inductive Hypothesis: Assume P(k) is true for some arbitrary integer k ≥ 1 (or k ≥ n₀). Write out what P(k) says explicitly — this is the assumption you will use.

Step 3 — Inductive Step: Using the assumption that P(k) is true, prove that P(k+1) must also be true. Start from P(k+1), relate it to P(k) using the inductive hypothesis, and complete the proof.

Standard sum formulas proved by Mathematical Induction (JEE Main maths — must know):

Σₙ = 1+2+3+…+n = n(n+1)/2

Σn² = 1²+2²+3²+…+n² = n(n+1)(2n+1)/6

Σn³ = 1³+2³+3³+…+n³ = [n(n+1)/2]² = (Σn)²

Σ(2n–1) = 1+3+5+…+(2n–1) = n² (sum of first n odd numbers)

Σ(2n) = 2+4+6+…+2n = n(n+1) (sum of first n even numbers)

a + ar + ar² + … + arⁿ⁻¹ = a(rⁿ–1)/(r–1) for r ≠ 1 (GP sum)

Standard divisibility results proved by Mathematical Induction (JEE Main):

n(n+1)(n+2) is divisible by 6 for all n ∈ ℕ (product of 3 consecutive integers)

n(n+1)(2n+1) is divisible by 6 for all n ∈ ℕ

2ⁿ > n for all n ∈ ℕ (proved by induction)

2ⁿ ≥ n² for all n ≥ 4 (proved by induction)

n! ≥ 2ⁿ⁻¹ for n ≥ 1

(1+x)ⁿ ≥ 1 + nx for x > –1 (Bernoulli's inequality)

4ⁿ – 1 is divisible by 3 for all n ∈ ℕ

7ⁿ – 1 is divisible by 6 for all n ∈ ℕ

n³ – n = n(n–1)(n+1) is divisible by 6 for all n ∈ ℕ

3^(2n+2) – 8n – 9 is divisible by 64 for all n ∈ ℕ (classic JEE induction problem)

10^n + 3·4^(n+2) + 5 is divisible by 9 for all n ∈ ℕ

Strong Induction (Complete Induction) in Mathematical Induction: Assumes P(1), P(2), …, P(k) are all true (not just P(k)) to prove P(k+1). Used when P(k+1) depends on more than just P(k). Download the Free PDF for Mathematical Induction and Binomial Theorem for all worked induction proofs.

Binomial Theorem — Expansion Formula and General Term

Why the General Term Formula of Binomial Theorem Is the Core JEE Main Formula

The Binomial Theorem states that for any positive integer n and any real or complex numbers a and b:

(a + b)ⁿ = ⁿC₀ aⁿ + ⁿC₁ aⁿ⁻¹b + ⁿC₂ aⁿ⁻²b² + … + ⁿCₙ bⁿ = Σᵣ₌₀ⁿ ⁿCᵣ aⁿ⁻ʳ bʳ

The Binomial Theorem expansion has (n+1) terms total. The coefficients ⁿC₀, ⁿC₁, …, ⁿCₙ are called binomial coefficients and are also written as C₀, C₁, …, Cₙ.

General Term of Binomial Theorem (Tᵣ₊₁) — JEE Main master formula:

Tᵣ₊₁ = ⁿCᵣ · aⁿ⁻ʳ · bʳ (the (r+1)ᵗʰ term in the expansion of (a+b)ⁿ)

Here r = 0, 1, 2, …, n. T₁ = ⁿC₀·aⁿ (first term); Tₙ₊₁ = ⁿCₙ·bⁿ (last term).

Applying the general term of Binomial Theorem in JEE Main:

For (x + 1/x)ⁿ: Tᵣ₊₁ = ⁿCᵣ · x^(n–r) · (1/x)ʳ = ⁿCᵣ · x^(n–2r). Set exponent = k to find coefficient of xᵏ → solve n–2r = k → r = (n–k)/2 (must be non-negative integer).

For (x + a)ⁿ: Tᵣ₊₁ = ⁿCᵣ · xⁿ⁻ʳ · aʳ. Coefficient of xᵏ = ⁿCₙ₋ₖ · aⁿ⁻ᵏ.

For (ax^p + b/x^q)ⁿ: Tᵣ₊₁ = ⁿCᵣ · aⁿ⁻ʳ · bʳ · x^(p(n–r)–qr). Set exponent = 0 (or k) and solve for r.

Sum of all coefficients of Binomial Theorem: put a = b = 1 → 2ⁿ. Put a = 1, b = –1 → 0 = sum of alternate terms.

Standard expansions from Binomial Theorem (JEE Main shortcuts):

(a–b)ⁿ = Σᵣ₌₀ⁿ (–1)ʳ ⁿCᵣ aⁿ⁻ʳ bʳ (alternate signs, first term positive)

(1+x)ⁿ = ⁿC₀ + ⁿC₁x + ⁿC₂x² + … + ⁿCₙxⁿ

(1–x)ⁿ = ⁿC₀ – ⁿC₁x + ⁿC₂x² – … + (–1)ⁿ ⁿCₙxⁿ

(a+b)ⁿ + (a–b)ⁿ = 2[ⁿC₀aⁿ + ⁿC₂aⁿ⁻²b² + ⁿC₄aⁿ⁻⁴b⁴ + …] (even terms)

(a+b)ⁿ – (a–b)ⁿ = 2[ⁿC₁aⁿ⁻¹b + ⁿC₃aⁿ⁻³b³ + …] (odd terms)

Download the Free PDF for Mathematical Induction and Binomial Theorem for the complete Binomial Theorem expansion table with worked general term examples for JEE Main.

Middle Term, Term Independent of x, and Greatest Term in Binomial Theorem

Why These Are the Most Directly Tested Binomial Theorem JEE Main Question Types

Middle term(s) of Binomial Theorem expansion (a+b)ⁿ:

When n is EVEN: one middle term → T₍ₙ/₂₎₊₁ = T₍ₙ₊₂₎/₂ (the single middle term at position (n/2)+1).

When n is ODD: two middle terms → T₍ₙ₊₁₎/₂ and T₍ₙ₊₃₎/₂ (the (n+1)/2 th and (n+3)/2 th terms). These two middle terms have equal binomial coefficients since ⁿC₍ₙ₋₁₎/₂ = ⁿC₍ₙ₊₁₎/₂.

Term independent of x in Binomial Theorem (JEE Main formula): In (ax^p + b·x^q)ⁿ, the general term is Tᵣ₊₁ = ⁿCᵣ·aⁿ⁻ʳ·bʳ·x^(p(n–r)+qr). Set the power of x = 0: p(n–r) + qr = 0 → pn – pr + qr = 0 → r(q–p) = –pn → r = pn/(p–q). This r must be a non-negative integer for the term independent of x to exist. If r is not an integer → no term independent of x exists in that Binomial Theorem expansion.

Numerically greatest term of Binomial Theorem (JEE Main formula): To find the greatest term in (1+x)ⁿ or (a+b)ⁿ:

Consider the ratio |Tᵣ₊₁/Tᵣ| = |(n–r+1)/r| · |b/a|. Set this ratio ≥ 1 to find when terms are increasing. The greatest term occurs at the value of r where this ratio transitions from ≥ 1 to < 1.

Condition: Tᵣ₊₁ ≥ Tᵣ when r ≤ (n+1)|b/a| / (1 + |b/a|) = m (say). If m is not an integer → greatest term is T₍⌊m⌋₊₁₎. If m is an integer → two equal greatest terms Tₘ and Tₘ₊₁. For (1+x)ⁿ: greatest term when r ≤ (n+1)x/(1+x). For (a+b)ⁿ: greatest term when r ≤ (n+1)|b/a| / (1+|b/a|).

Greatest binomial coefficient term (Binomial Theorem): Among all binomial coefficients ⁿC₀, ⁿC₁, …, ⁿCₙ, the greatest is ⁿCₙ/₂ (n even) or ⁿC₍ₙ₋₁₎/₂ = ⁿC₍ₙ₊₁₎/₂ (n odd). Note: greatest binomial coefficient ≠ greatest term (greatest term also depends on a and b values). Download the Free PDF for Binomial Theorem middle term and greatest term worked examples for JEE Main.

Binomial Coefficients — Properties and Sum Identities for JEE Main

Why Binomial Coefficient Identities Are the Most Formula-Dense Binomial Theorem Section

Binomial coefficients C₀, C₁, C₂, …, Cₙ (shorthand for ⁿC₀, ⁿC₁, …, ⁿCₙ) have a rich family of identities in the Binomial Theorem — all provable by substituting specific values of x into (1+x)ⁿ = C₀+C₁x+C₂x²+…+Cₙxⁿ, or by differentiating, integrating, or multiplying two expansions.

Core binomial coefficient identities in Binomial Theorem (JEE Main formulas — all must be memorised):

(i) C₀ + C₁ + C₂ + … + Cₙ = 2ⁿ — put x = 1 in (1+x)ⁿ

(ii) C₀ – C₁ + C₂ – C₃ + … + (–1)ⁿCₙ = 0 — put x = –1 in (1+x)ⁿ

(iii) C₀ + C₂ + C₄ + … = C₁ + C₃ + C₅ + … = 2ⁿ⁻¹ — add and subtract (i) and (ii)

(iv) ΣrCᵣ = n·2ⁿ⁻¹ — differentiate (1+x)ⁿ, put x = 1

(v) Σr(r–1)Cᵣ = n(n–1)·2ⁿ⁻² — differentiate twice, put x = 1

(vi) Σr²Cᵣ = n(n+1)·2ⁿ⁻² — from r²= r(r–1)+r, combine (iv) and (v)

(vii) ΣCᵣ/r = (2ⁿ⁺¹–1)/(n+1) — integrate (1+x)ⁿ from 0 to 1 → C₀+C₁/2+C₂/3+…+Cₙ/(n+1) = (2ⁿ⁺¹–1)/(n+1)

(viii) C₀/1 + C₁/2 + C₂/3 + … + Cₙ/(n+1) = (2ⁿ⁺¹–1)/(n+1)

(ix) C₀C₁ + C₁C₂ + C₂C₃ + … + Cₙ₋₁Cₙ = ²ⁿCₙ₋₁ — from Vandermonde's identity

(x) C₀²+C₁²+C₂²+…+Cₙ² = ²ⁿCₙ — coefficient of xⁿ in (1+x)ⁿ·(1+x)ⁿ = (1+x)²ⁿ; also written as Σ(Cᵣ)² = ²ⁿCₙ

(xi) C₀Cᵣ + C₁Cᵣ₊₁ + C₂Cᵣ₊₂ + … + Cₙ₋ᵣCₙ = ²ⁿCₙ₋ᵣ = ²ⁿCₙ₊ᵣ (Vandermonde's identity)

(xii) C₀C₁ + C₂C₃ + … = ²ⁿ⁻¹Cₙ₋₁ + ²ⁿ⁻¹Cₙ (mixed product sums)

(xiii) ΣCᵣ·Cₛ (sum over all r+s=n) = ²ⁿCₙ (coeff. of xⁿ in (1+x)²ⁿ)

(xiv) 3C₀ + 3²C₁/2 + 3³C₂/3 + … + 3ⁿ⁺¹Cₙ/(n+1) = (4ⁿ⁺¹–1)/(3(n+1)) — integrate (1+3x)ⁿ from 0 to 1

Download the Free PDF for Mathematical Induction and Binomial Theorem for all binomial coefficient identity proofs and applications for JEE Main.

Binomial Theorem for Negative and Fractional Index — Infinite Series Expansion

Why the Infinite Binomial Expansion Is a JEE Main Approximation and Coefficient Formula

The Binomial Theorem extends to non-integer (negative, rational, fractional) exponents when |x| < 1. For |x| < 1 and any real n (not necessarily a positive integer):

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

Standard infinite binomial expansions (JEE Main formulas):

(1+x)⁻¹ = 1 – x + x² – x³ + x⁴ – … (|x|<1) — GP with ratio –x

(1–x)⁻¹ = 1 + x + x² + x³ + x⁴ + … (|x|<1) — GP with ratio x

(1+x)⁻² = 1 – 2x + 3x² – 4x³ + … (|x|<1)

(1–x)⁻² = 1 + 2x + 3x² + 4x³ + … (|x|<1)

(1+x)⁻³ = 1 – 3x + 6x² – 10x³ + … = Σ(–1)ʳ ⁽ʳ⁺²⁾C₂ xʳ

(1–x)⁻³ = 1 + 3x + 6x² + 10x³ + … = Σ ⁽ʳ⁺²⁾C₂ xʳ

(1+x)^(1/2) = 1 + x/2 – x²/8 + x³/16 – … (|x|<1, approximation of √(1+x))

(1+x)^(–1/2) = 1 – x/2 + 3x²/8 – 5x³/16 + … (approximation of 1/√(1+x))

General term for (1+x)ⁿ infinite expansion (Binomial Theorem for non-integer n):

Tᵣ₊₁ = n(n–1)(n–2)…(n–r+1) · xʳ / r! (r = 0, 1, 2, …, ∞)

Note: for non-integer n, the series is infinite (not finite like positive integer case) and converges only for |x| < 1.

Approximations using Binomial Theorem (JEE Main applications): When x is very small (|x| << 1), higher-order terms are negligible: (1+x)ⁿ ≈ 1 + nx. (1–x)ⁿ ≈ 1 – nx. √(1+x) ≈ 1 + x/2. 1/√(1+x) ≈ 1 – x/2. These approximations appear in physics numerical questions and JEE Main maths approximation problems. Download the Free PDF for Mathematical Induction and Binomial Theorem for all infinite expansion examples.

Remainder Theorem and Divisibility Using Binomial Theorem

Why Remainder and Divisibility Problems Are Classic JEE Main Binomial Theorem Applications

One of the most elegant applications of the Binomial Theorem in JEE Main is finding the remainder when expressions like Nⁿ are divided by a number, by writing N as (m ± 1) or (m ± r) and expanding using the Binomial Theorem.

Remainder technique using Binomial Theorem (JEE Main):

Method 1: Write the base as (1 + k·p) or (k·p – 1) where p is the divisor. Expand using Binomial Theorem → all terms except the last are divisible by p → remainder = last term.

Example: Find remainder when 7¹⁰⁰ is divided by 6. Write 7 = 6+1: 7¹⁰⁰ = (6+1)¹⁰⁰ = Σ¹⁰⁰Cᵣ·6ʳ·1¹⁰⁰⁻ʳ. All terms with r ≥ 1 are divisible by 6. Remainder = ¹⁰⁰C₀·1¹⁰⁰ = 1.

Example: Find remainder when 3²³ is divided by 4. Write 3 = 4–1: 3²³ = (4–1)²³ = Σ²³Cᵣ·4²³⁻ʳ·(–1)ʳ. Only the last term (r=23) is not divisible by 4: (–1)²³ = –1 ≡ 3 (mod 4). Remainder = 3.

Binomial Theorem — Integer and fractional part problems (JEE Main): For expressions of the form (p + √q)ⁿ + (p – √q)ⁿ where p, q, n are positive integers: since (p – √q) and (p + √q) are conjugates, their sum eliminates the irrational parts → the result is an even integer. Key result: if I = integer part and f = fractional part of (p+√q)ⁿ, and f' = fractional part of (p–√q)ⁿ, then f + f' = 1 (if 0 < p–√q < 1) and (I+f)·f = (p²–q)ⁿ. Classic JEE problem: if (√3+1)²ⁿ = I + f, then I is odd and f(I+1) = 2²ⁿ. Download the Free PDF for Mathematical Induction and Binomial Theorem for remainder and integer-fractional part problems.

Multinomial Theorem and Number of Terms in Expansions

Why Multinomial Theorem Extends Binomial Theorem for JEE Main Multi-Variable Problems

The Multinomial Theorem extends the Binomial Theorem to expansions of (a₁ + a₂ + … + aₘ)ⁿ where m variables are involved. The general term is:

General term of Multinomial Theorem: n! / (n₁! · n₂! · … · nₘ!) · a₁^n₁ · a₂^n₂ · … · aₘ^nₘ where n₁+n₂+…+nₘ = n (and each nᵢ ≥ 0).

The multinomial coefficient n!/(n₁!n₂!…nₘ!) is the number of ways to arrange n objects where n₁ are of type 1, n₂ of type 2, etc.

Number of terms in Multinomial Theorem expansion: The expansion of (a₁+a₂+…+aₘ)ⁿ has ⁿ⁺ᵐ⁻¹Cₘ₋₁ distinct terms (stars and bars for n₁+n₂+…+nₘ = n, nᵢ ≥ 0).

Number of terms in standard expansions (Binomial Theorem — JEE Main):

(a+b+c)ⁿ: number of terms = ⁿ⁺²C₂ = (n+1)(n+2)/2

(a+b+c+d)ⁿ: number of terms = ⁿ⁺³C₃ = (n+1)(n+2)(n+3)/6

(a₁+a₂+…+aₘ)ⁿ: number of terms = ⁿ⁺ᵐ⁻¹Cₘ₋₁

Coefficient extraction using Multinomial Theorem (JEE Main): Coefficient of x^p · y^q · z^r in (x+y+z)ⁿ where p+q+r=n = n!/(p!q!r!). Coefficient of x^p · y^q in (x+y)^(p+q) = (p+q)!/(p!q!) = C(p+q,p). Sum of all coefficients of (x+y+z)ⁿ: put x=y=z=1 → 3ⁿ. Download the Free PDF for Multinomial Theorem terms and coefficient formulas for JEE Main.

Key Standard Results and Applications of Binomial Theorem for JEE Main

Why These Binomial Theorem Results Are Guaranteed JEE Main Formula Questions

Sum of coefficients trick in Binomial Theorem (JEE Main shortcut): To find sum of all coefficients in expansion of f(x) = (polynomial in x)ⁿ, simply put x = 1 in f(x). Example: sum of coefficients of (2x+3y)⁸ = put x=y=1 → 5⁸. Sum of coefficients of (1+2x–3x²)⁷ = put x=1 → (1+2–3)⁷ = 0⁷ = 0.

Binomial Theorem for finding the last digit (unit digit): Express the base as sum of a multiple of 10 and a small number, then use Binomial Theorem to find unit digit of the last term only. Example: unit digit of 2⁴⁰⁰ — since 2⁴=16 (unit digit 6), and 6ⁿ always ends in 6 → unit digit of 2⁴⁰⁰ = (2⁴)¹⁰⁰ = 16¹⁰⁰ → ends in 6. Unit digit of 7ⁿ: cycles with period 4 → 7¹=7, 7²=9, 7³=3, 7⁴=1, then repeats.

Binomial Theorem — ratio of consecutive terms: |Tᵣ₊₁/Tᵣ| = |(n–r+1)/r| · |b/a|. Using this to find when terms increase or decrease → determines the greatest term position.

Finding coefficient of xⁿ in product of two series using Binomial Theorem: If the coefficient of xⁿ is needed in (1+x)ᵃ·(1+x)ᵇ = (1+x)^(a+b): it is ^(a+b)Cₙ. More generally: coefficient of xⁿ in Σⱼ₌₀ᵃ ᵃCⱼ·xʲ · Σₖ₌₀ᵇ ᵇCₖ·xᵏ = Σⱼ₊ₖ₌ₙ ᵃCⱼ·ᵇCₙ₋ⱼ = ^(a+b)Cₙ (Vandermonde's convolution). This identity connects Binomial Theorem coefficients to combination counting.

Binomial Theorem — rationalising and simplifying expressions: (√x + √y)ⁿ + (√x – √y)ⁿ = 2[xⁿ/² + ⁿC₂xⁿ/²⁻¹y + ⁿC₄xⁿ/²⁻²y² + …] (even terms only, rational if x, y rational). (√x + √y)ⁿ – (√x – √y)ⁿ = 2√y[ⁿC₁xⁿ/²⁻¹/²y⁰ + ⁿC₃xⁿ/²⁻³/²y + …] (odd terms, irrational part). Download the Free PDF for Mathematical Induction and Binomial Theorem for all standard application results.