JEE MATH APEX
JEE MATH APEX
JEE MATH APEX
JEE MATH APEX
JEE MATH APEX
JEE MATH APEX
JEE MATH APEX
JEE MATH APEX
JEE MATH APEX
JEE MATH APEX
BINOMIAL THEOREM
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JEE MATH APEX
1. Binomial Theorem for Positive Integer
📌 Expansion Formula
• (x + y)ⁿ = ⁿC₀xⁿ + ⁿC₁xⁿ⁻¹y + ⁿC₂xⁿ⁻²y² + … + ⁿCₙyⁿ
• (x + y)ⁿ = Σr=0ⁿ ⁿCᵣ xⁿ⁻ʳ yʳ
• General term: Tr+1 = ⁿCᵣ xⁿ⁻ʳ yʳ
• Number of terms: n + 1
• (1 + x)ⁿ = ⁿC₀ + ⁿC₁x + ⁿC₂x² + … + ⁿCₙxⁿ
• (x + y)ⁿ = Σr=0ⁿ ⁿCᵣ xⁿ⁻ʳ yʳ
• General term: Tr+1 = ⁿCᵣ xⁿ⁻ʳ yʳ
• Number of terms: n + 1
• (1 + x)ⁿ = ⁿC₀ + ⁿC₁x + ⁿC₂x² + … + ⁿCₙxⁿ
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2. General Term & Middle Term
📌 Finding Specific Terms
• General term (r+1): Tr+1 = ⁿCᵣ xⁿ⁻ʳ yʳ
• Coefficient of xᵏ in (x + y)ⁿ: ⁿC₍ₙ₋ₖ₎ or ⁿCₖ
• Middle term when n is even: T(n/2)+1
• Middle terms when n is odd: T(n+1)/2 and T(n+3)/2
• Term independent of x: Put power of x = 0
• Coefficient of xᵏ in (x + y)ⁿ: ⁿC₍ₙ₋ₖ₎ or ⁿCₖ
• Middle term when n is even: T(n/2)+1
• Middle terms when n is odd: T(n+1)/2 and T(n+3)/2
• Term independent of x: Put power of x = 0
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3. Properties of Binomial Coefficients
📌 Sum of Coefficients
• ⁿC₀ + ⁿC₁ + ⁿC₂ + … + ⁿCₙ = 2ⁿ
• ⁿC₀ + ⁿC₂ + ⁿC₄ + … = 2ⁿ⁻¹
• ⁿC₁ + ⁿC₃ + ⁿC₅ + … = 2ⁿ⁻¹
• ⁿC₀ + ⁿC₁·x + ⁿC₂·x² + … + ⁿCₙ·xⁿ = (1 + x)ⁿ
• Sum of coefficients in (1 + x)ⁿ = 2ⁿ (put x = 1)
• ⁿC₀ + ⁿC₂ + ⁿC₄ + … = 2ⁿ⁻¹
• ⁿC₁ + ⁿC₃ + ⁿC₅ + … = 2ⁿ⁻¹
• ⁿC₀ + ⁿC₁·x + ⁿC₂·x² + … + ⁿCₙ·xⁿ = (1 + x)ⁿ
• Sum of coefficients in (1 + x)ⁿ = 2ⁿ (put x = 1)
JEE MATH APEX
4. Sum of Coefficients (Special Cases)
📌 Sum of Coefficients in Different Expansions
• Sum of coefficients in (1 + x)ⁿ: 2ⁿ (put x = 1)
• Sum of coefficients in (1 − x)ⁿ: 0 (put x = 1)
• Sum of coefficients in (a + bx)ⁿ: (a + b)ⁿ (put x = 1)
• Sum of even coefficients = Sum of odd coefficients = 2ⁿ⁻¹
• Sum of coefficients in (1 + x + x²)ⁿ: 3ⁿ
• Sum of coefficients in (1 − x)ⁿ: 0 (put x = 1)
• Sum of coefficients in (a + bx)ⁿ: (a + b)ⁿ (put x = 1)
• Sum of even coefficients = Sum of odd coefficients = 2ⁿ⁻¹
• Sum of coefficients in (1 + x + x²)ⁿ: 3ⁿ
JEE MATH APEX
5. Identities Involving Binomial Coefficients
📌 Important Identities
• ⁿC₀·x + ⁿC₁·x²/2 + ⁿC₂·x³/3 + … = [(1+x)ⁿ⁺¹ − 1]/(n+1)
• ⁿC₀ − ⁿC₁ + ⁿC₂ − ⁿC₃ + … + (−1)ⁿ ⁿCₙ = 0
• ⁿC₀·r + ⁿC₁·(r+1) + … = 2ⁿ⁻¹(n + 2r)
• ⁿC₀² + ⁿC₁² + ⁿC₂² + … + ⁿCₙ² = ²ⁿCₙ
• ⁿC₀·ᵐCᵣ + ⁿC₁·ᵐCᵣ₋₁ + … = ⁿ⁺ᵐCᵣ (Vandermonde’s Identity)
• ⁿC₀ − ⁿC₁ + ⁿC₂ − ⁿC₃ + … + (−1)ⁿ ⁿCₙ = 0
• ⁿC₀·r + ⁿC₁·(r+1) + … = 2ⁿ⁻¹(n + 2r)
• ⁿC₀² + ⁿC₁² + ⁿC₂² + … + ⁿCₙ² = ²ⁿCₙ
• ⁿC₀·ᵐCᵣ + ⁿC₁·ᵐCᵣ₋₁ + … = ⁿ⁺ᵐCᵣ (Vandermonde’s Identity)
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6. Greatest Term
📌 Numerically Greatest Term in (1 + x)ⁿ
• In (1 + x)ⁿ, Tr+1 and Tr compared: Tr+1/Tr = (n−r+1)/r · |x|
• Greatest term occurs when: (n+1)|x|/(|x|+1) ≤ r ≤ (n+1)|x|/(|x|+1) + 1
• If (n+1)|x|/(|x|+1) is integer m, then Tm = Tm+1 (two greatest terms)
• Numerically greatest term in (a + bx)ⁿ: Find r for |Tr+1| maximum
• Greatest term occurs when: (n+1)|x|/(|x|+1) ≤ r ≤ (n+1)|x|/(|x|+1) + 1
• If (n+1)|x|/(|x|+1) is integer m, then Tm = Tm+1 (two greatest terms)
• Numerically greatest term in (a + bx)ⁿ: Find r for |Tr+1| maximum
JEE MATH APEX
7. Binomial Theorem for Any Index
📌 For Negative or Fractional Index
• (1 + x)ⁿ = 1 + nx + [n(n−1)/2!]x² + [n(n−1)(n−2)/3!]x³ + …
• Valid for |x| < 1
• General term: Tr+1 = [n(n−1)(n−2)…(n−r+1)/r!] · xʳ
• (1 + x)⁻¹ = 1 − x + x² − x³ + … (|x| < 1)
• (1 − x)⁻¹ = 1 + x + x² + x³ + … (|x| < 1)
• Valid for |x| < 1
• General term: Tr+1 = [n(n−1)(n−2)…(n−r+1)/r!] · xʳ
• (1 + x)⁻¹ = 1 − x + x² − x³ + … (|x| < 1)
• (1 − x)⁻¹ = 1 + x + x² + x³ + … (|x| < 1)
JEE MATH APEX
8. Binomial Coefficients with Products
📌 Series Involving Products
• ⁿC₀·ᵐC₀ + ⁿC₁·ᵐC₁ + … + ⁿCᵣ·ᵐCᵣ = ⁿ⁺ᵐCᵣ (Vandermonde)
• (ⁿC₀)² + (ⁿC₁)² + … + (ⁿCₙ)² = ²ⁿCₙ
• ⁿC₀·ⁿC₁ + ⁿC₁·ⁿC₂ + … + ⁿCₙ₋₁·ⁿCₙ = ²ⁿCₙ₊₁
• ⁿC₀ − ⁿC₂ + ⁿC₄ − ⁿC₆ + … = 2ⁿᐟ² cos(nπ/4)
• ⁿC₁ − ⁿC₃ + ⁿC₅ − ⁿC₇ + … = 2ⁿᐟ² sin(nπ/4)
• (ⁿC₀)² + (ⁿC₁)² + … + (ⁿCₙ)² = ²ⁿCₙ
• ⁿC₀·ⁿC₁ + ⁿC₁·ⁿC₂ + … + ⁿCₙ₋₁·ⁿCₙ = ²ⁿCₙ₊₁
• ⁿC₀ − ⁿC₂ + ⁿC₄ − ⁿC₆ + … = 2ⁿᐟ² cos(nπ/4)
• ⁿC₁ − ⁿC₃ + ⁿC₅ − ⁿC₇ + … = 2ⁿᐟ² sin(nπ/4)
JEE MATH APEX
9. Divisibility Problems
📌 Using Binomial Theorem for Divisibility
• To prove (1 + x)ⁿ = 1 + nx + multiple of x²
• (1 + x)ⁿ − 1 − nx is divisible by x²
• (1 + x)ⁿ − 1 is divisible by x
• (1 + x)ⁿ = 1 + nx + ⁿC₂x² + … + xⁿ
• (a + b)ⁿ = aⁿ + ⁿC₁aⁿ⁻¹b + … (for divisibility)
• (1 + x)ⁿ − 1 − nx is divisible by x²
• (1 + x)ⁿ − 1 is divisible by x
• (1 + x)ⁿ = 1 + nx + ⁿC₂x² + … + xⁿ
• (a + b)ⁿ = aⁿ + ⁿC₁aⁿ⁻¹b + … (for divisibility)
JEE MATH APEX
10. Multinomial Expansion
📌 (x + y + z)ⁿ Expansion
• (x + y + z)ⁿ = Σ [n!/(p!q!r!)] · xᵖyᵠzʳ
• Where p + q + r = n
• Number of terms: (n+1)(n+2)/2
• Coefficient of xᵖyᵠzʳ: n!/(p!q!r!)
• Sum of all coefficients: 3ⁿ
• Where p + q + r = n
• Number of terms: (n+1)(n+2)/2
• Coefficient of xᵖyᵠzʳ: n!/(p!q!r!)
• Sum of all coefficients: 3ⁿ
JEE MATH APEX
11. Binomial Coefficients with Powers
📌 Sum of Products with Natural Numbers
• 1·ⁿC₁ + 2·ⁿC₂ + 3·ⁿC₃ + … + n·ⁿCₙ = n·2ⁿ⁻¹
• 1²·ⁿC₁ + 2²·ⁿC₂ + 3²·ⁿC₃ + … = n(n+1)·2ⁿ⁻²
• ⁿC₀/1 + ⁿC₁/2 + ⁿC₂/3 + … + ⁿCₙ/(n+1) = (2ⁿ⁺¹ − 1)/(n+1)
• ⁿC₀ − ⁿC₁/2 + ⁿC₂/3 − ⁿC₃/4 + … = 1/(n+1)
• 1²·ⁿC₁ + 2²·ⁿC₂ + 3²·ⁿC₃ + … = n(n+1)·2ⁿ⁻²
• ⁿC₀/1 + ⁿC₁/2 + ⁿC₂/3 + … + ⁿCₙ/(n+1) = (2ⁿ⁺¹ − 1)/(n+1)
• ⁿC₀ − ⁿC₁/2 + ⁿC₂/3 − ⁿC₃/4 + … = 1/(n+1)
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12. Applications in Approximation
📌 Using Binomial for Approximations
• (1 + x)ⁿ ≈ 1 + nx (when |x| is small)
• (1 + x)ⁿ ≈ 1 + nx + [n(n−1)/2]x²
• √(1 + x) ≈ 1 + x/2 − x²/8 + x³/16
• 1/(1 + x) ≈ 1 − x + x² − x³
• 1/(1 − x) ≈ 1 + x + x² + x³
• (1 + x)ⁿ ≈ 1 + nx + [n(n−1)/2]x²
• √(1 + x) ≈ 1 + x/2 − x²/8 + x³/16
• 1/(1 + x) ≈ 1 − x + x² − x³
• 1/(1 − x) ≈ 1 + x + x² + x³
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13. Remainder & Last Digit Problems
📌 Finding Remainders Using Binomial
• To find remainder of (a + b)ⁿ divided by m:
Write a = km + r, then use binomial
• Last digit of aⁿ: Find a mod 10
• Remainder of (1 + x)ⁿ divided by x² is 1 + nx
• Remainder of (1 + x)ⁿ divided by x is 1
Write a = km + r, then use binomial
• Last digit of aⁿ: Find a mod 10
• Remainder of (1 + x)ⁿ divided by x² is 1 + nx
• Remainder of (1 + x)ⁿ divided by x is 1
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14. Coefficient Problems (Advanced)
📌 Finding Coefficients in Products
• Coefficient of xʳ in (1 + x)ᵐ(1 + x)ⁿ = (1 + x)ᵐ⁺ⁿ: ᵐ⁺ⁿCᵣ
• Coefficient of xʳ in (1 + x)ᵐ(1 − x)ⁿ: Use expansion and combine
• Coefficient of xⁿ in (1 + x)²ⁿ: ²ⁿCₙ
• Coefficient of xʳ in (1 + x + x²)ⁿ: Σ [n!/(p!q!r!)] where p+2q = r
• Coefficient of xʳ in (1 + x)ᵐ(1 − x)ⁿ: Use expansion and combine
• Coefficient of xⁿ in (1 + x)²ⁿ: ²ⁿCₙ
• Coefficient of xʳ in (1 + x + x²)ⁿ: Σ [n!/(p!q!r!)] where p+2q = r
✅ BINOMIAL THEOREM ✅
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— JEE MATH APEX —
JEE MATH APEX 
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