📚 Basic Binomial Theorem Formulas
1. Binomial Theorem (Positive Integer Index):
(x + y)ⁿ = ⁿC₀xⁿy⁰ + ⁿC₁xⁿ⁻¹y¹ + ⁿC₂xⁿ⁻²y² + ... + ⁿCₙx⁰yⁿ= Σ ⁿCᵣ xⁿ⁻ʳ yʳ (r = 0 to n)
2. General Term (rth term from beginning):
Tᵣ = ⁿCᵣ₋₁ xⁿ⁻⁽ʳ⁻¹⁾ yʳ⁻¹Or: Tᵣ₊₁ = ⁿCᵣ xⁿ⁻ʳ yʳ
3. Number of Terms:
Total terms in expansion of (x + y)ⁿ = n + 1
4. Sum of Binomial Coefficients:
ⁿC₀ + ⁿC₁ + ⁿC₂ + ... + ⁿCₙ = 2ⁿ
5. Sum of Binomial Coefficients at Odd Places:
ⁿC₀ + ⁿC₂ + ⁿC₄ + ... = 2ⁿ⁻¹
6. Sum of Binomial Coefficients at Even Places:
ⁿC₁ + ⁿC₃ + ⁿC₅ + ... = 2ⁿ⁻¹
Remember: Put x = y = 1 to get sum of all coefficients
Put x = 1, y = -1 to prove odd = even places
Put x = 1, y = -1 to prove odd = even places
⭐ Important Properties
7. Symmetric Property:
ⁿCᵣ = ⁿCₙ₋ᵣ
8. Pascal's Identity:
ⁿCᵣ + ⁿCᵣ₋₁ = ⁿ⁺¹Cᵣ
9. Ratio of Consecutive Terms:
Tᵣ₊₁/Tᵣ = [(n-r+1)/r] × (y/x)
10. Coefficient of xʳ in (1+x)ⁿ:
Coefficient of xʳ = ⁿCᵣ
11. rth Term from End:
rth term from end = (n-r+2)th term from beginning
Quick Trick: For (x + a)ⁿ, replace a with variable and find power!
🎯 Middle Terms
12. When n is Even:
Middle term = [(n/2) + 1]th termOnly ONE middle term = T₍ₙ/₂₎₊₁
13. When n is Odd:
Two middle terms = [(n+1)/2]th and [(n+3)/2]th termsT₍ₙ₊₁₎/₂ and T₍ₙ₊₃₎/₂
Example:
• (x + y)⁶: n = 6 (even) → Middle term = T₄
• (x + y)⁷: n = 7 (odd) → Middle terms = T₄ and T₅
• (x + y)⁶: n = 6 (even) → Middle term = T₄
• (x + y)⁷: n = 7 (odd) → Middle terms = T₄ and T₅
📊 Greatest Term & Greatest Coefficient
14. Greatest Coefficient:
• If n is even: ⁿCₙ/₂• If n is odd: ⁿC₍ₙ₋₁₎/₂ = ⁿC₍ₙ₊₁₎/₂
15. Greatest Term in (x + y)ⁿ:
Find m such that: [(n+1)|y|] / [|x| + |y|] ≤ m < [(n+1)|y|] / [|x| + |y|] + 1Greatest term = Tₘ or Tₘ₊₁
16. Greatest Term in (1 + x)ⁿ:
m = [(n+1)x] / [1+x]• If m is integer: Two greatest terms Tₘ and Tₘ₊₁
• If m is not integer: Greatest term = T₍ₘ₊₁₎ where [m] = greatest integer ≤ m
🌟 Special Expansions & Results
17. (1 + x)ⁿ:
= ⁿC₀ + ⁿC₁x + ⁿC₂x² + ⁿC₃x³ + ... + ⁿCₙxⁿ
18. (1 - x)ⁿ:
= ⁿC₀ - ⁿC₁x + ⁿC₂x² - ⁿC₃x³ + ... + (-1)ⁿ ⁿCₙxⁿ
19. (x + 1/x)ⁿ:
= ⁿC₀xⁿ + ⁿC₁xⁿ⁻² + ⁿC₂xⁿ⁻⁴ + ... (Powers decrease by 2)
20. (x - 1/x)ⁿ:
= ⁿC₀xⁿ - ⁿC₁xⁿ⁻² + ⁿC₂xⁿ⁻⁴ - ... (Alternating signs)
21. (1 + x)ⁿ + (1 - x)ⁿ:
= 2[ⁿC₀ + ⁿC₂x² + ⁿC₄x⁴ + ...] (Only even powers)
22. (1 + x)ⁿ - (1 - x)ⁿ:
= 2[ⁿC₁x + ⁿC₃x³ + ⁿC₅x⁵ + ...] (Only odd powers)
🔢 Multinomial Theorem
23. Multinomial Theorem:
(x₁ + x₂ + x₃ + ... + xₖ)ⁿ = Σ [n!/(r₁!r₂!...rₖ!)] x₁ʳ¹x₂ʳ²...xₖʳᵏwhere r₁ + r₂ + ... + rₖ = n
24. Number of Terms in (x₁ + x₂ + ... + xₖ)ⁿ:
= ⁿ⁺ᵏ⁻¹Cₖ₋₁ = ⁿ⁺ᵏ⁻¹Cₙ
25. For (x + y + z)ⁿ:
Number of terms = ⁿ⁺²C₂ = (n+1)(n+2)/2
🎓 Advanced Results
26. C₀ + 2C₁ + 3C₂ + ... + (n+1)Cₙ:
= (n+2)·2ⁿ⁻¹
27. C₀ - C₁ + C₂ - C₃ + ... + (-1)ⁿCₙ:
= 0
28. C₀² + C₁² + C₂² + ... + Cₙ²:
= ²ⁿCₙ
29. C₀C₁ + C₁C₂ + C₂C₃ + ... + Cₙ₋₁Cₙ:
= ²ⁿCₙ₊₁
30. Coefficient of xⁿ in (1+x)(1+x²)(1+x⁴)...:
Always 0 or 1 (Each power can be chosen at most once)
For Exam: Differentiate (1+x)ⁿ and put x=1 to get sum formulas!
📋 Quick Formula Reference
| Formula Type | Formula |
|---|---|
| General Term | Tᵣ₊₁ = ⁿCᵣ xⁿ⁻ʳ yʳ |
| Middle (n even) | T₍ₙ/₂₎₊₁ |
| Middle (n odd) | T₍ₙ₊₁₎/₂ and T₍ₙ₊₃₎/₂ |
| Sum of Coefficients | 2ⁿ |
| Greatest Coefficient (n even) | ⁿCₙ/₂ |
| Greatest Coefficient (n odd) | ⁿC₍ₙ₋₁₎/₂ = ⁿC₍ₙ₊₁₎/₂ |
| Term from End | rth from end = (n-r+2)th from beginning |
📝 EAPCET 2018-2025 - Real Previous Year Questions
TS EAMCET 2018 Mathematics Questions
A collection of binomial theorem and infinite series problems with beautiful LaTeX rendering
Question1. The absolute value of the numerically greatest term in the expansion of \((2x - 3y)^{12}\) when \(x = 3, y = 2\) is:
Question2. The sum to infinite terms of the series \[ x = \frac{3\}{10} + \frac{2\cdot 5}{3\cdot 6} + \frac{2\cdot 5\cdot 8}{3\cdot 6\cdot 9} + \frac{2\cdot 5\cdot 8\cdot 11}{3\cdot 6\cdot 9\cdot 12} + \cdots, \] is:
Q3. If \(n\) is a positive integer and the coefficient of \(x^{10}\) in the expansion of \((1+x)^{15}\) is equal to the coefficient of \(x^5\) in the expansion of \((1-x)^n\), then \(n=\)
Q4. If \[ x = \frac{2\cdot 5}{3\cdot 6} + \frac{2\cdot 5\cdot 8}{3\cdot 6\cdot 9} + \frac{2\cdot 5\cdot 8\cdot 11}{3\cdot 6\cdot 9\cdot 12} + \cdots, \] then evaluate \(72\,(12x+55)^3\).
Q17. If the coefficients of the terms \((2x+4)^n\) and \((x-2)^n\) in the expansion of \((1+x)^{2019}\) are equal, then \(\alpha =\)
Q18. If \(n\) is a positive integer, then the coefficient of \(x^6\) in the expansion of \[ (1 - 2x + 3x^2 - 4x^3 + \cdots)^n \] is:
TS EAMCET 2018 — 7th May Shift 1
The middle term in the expansion of \((2x - \tfrac{3}{x})^6\) is:
The sum of the binomial coefficients in the expansion of \((1+x)^5\) is:
Q7. If the coefficients of the \(r^{\text{th}}, (r+1)^{\text{th}}, (r+2)^{\text{th}}\) terms in the expansion of \((1+x)^n\) are in the ratio \(2:4:5\), then the values of \(r, n\) are:
Q8. If \(n\) is a positive integer greater than 1, then evaluate \[ (\,^nC_0)^{-8} (\,^nC_1)^{13} (\,^nC_2)^{-18} (\,^nC_3)^{25} \cdots \text{ up to } (n+1)\text{ terms}. \]
Q1. Find the coefficient of x⁵ in the expansion of (1 + 2x)⁶.
Answer: 192
Solution: T₆ = ⁶C₅(1)¹(2x)⁵ = 6 × 32x⁵ = 192x⁵. Coefficient = 192.
Solution: T₆ = ⁶C₅(1)¹(2x)⁵ = 6 × 32x⁵ = 192x⁵. Coefficient = 192.
Q2. In the expansion of (x + 1/x)¹⁰, find the term independent of x.
Answer: 252
Solution: For term independent of x: x¹⁰⁻ʳ · x⁻ʳ = x⁰, so 10-2r = 0, r = 5. T₆ = ¹⁰C₅ = 252.
Solution: For term independent of x: x¹⁰⁻ʳ · x⁻ʳ = x⁰, so 10-2r = 0, r = 5. T₆ = ¹⁰C₅ = 252.
Q3. The middle term in the expansion of (2x - 3y)⁸ is:
Answer: 489888x⁴y⁴
Solution: n = 8 (even), middle term = T₅ = ⁸C₄(2x)⁴(-3y)⁴ = 70 × 16x⁴ × 81y⁴ = 90720x⁴y⁴.
Solution: n = 8 (even), middle term = T₅ = ⁸C₄(2x)⁴(-3y)⁴ = 70 × 16x⁴ × 81y⁴ = 90720x⁴y⁴.
Q4. Find the sum of the binomial coefficients in the expansion of (x + y)¹⁰.
Answer: 1024
Solution: Sum of coefficients = 2ⁿ = 2¹⁰ = 1024.
Solution: Sum of coefficients = 2ⁿ = 2¹⁰ = 1024.
Q5. If the coefficient of x⁷ and x⁸ in (2 + x/3)ⁿ are equal, find n.
Answer: 55
Solution: ⁿC₇(2)ⁿ⁻⁷(1/3)⁷ = ⁿC₈(2)ⁿ⁻⁸(1/3)⁸. Simplifying: (n-7)/(8) = 1/(3×2) gives n = 55.
Solution: ⁿC₇(2)ⁿ⁻⁷(1/3)⁷ = ⁿC₈(2)ⁿ⁻⁸(1/3)⁸. Simplifying: (n-7)/(8) = 1/(3×2) gives n = 55.
Q6. The ratio of the 5th term from the beginning to the 5th term from the end in (√2 + 1/√3)¹⁰ is:
Answer: 1:16
Solution: T₅ = ¹⁰C₄(√2)⁶(1/√3)⁴. 5th from end = T₇ = ¹⁰C₆(√2)⁴(1/√3)⁶. Ratio = (2³/3²)/(2²/3³) = 6/96 = 1/16.
Solution: T₅ = ¹⁰C₄(√2)⁶(1/√3)⁴. 5th from end = T₇ = ¹⁰C₆(√2)⁴(1/√3)⁶. Ratio = (2³/3²)/(2²/3³) = 6/96 = 1/16.
Q7. Find the term containing x⁶ in the expansion of (x² - 2/x)¹⁰.
Answer: 3360x⁶
Solution: Tᵣ₊₁ = ¹⁰Cᵣ(x²)¹⁰⁻ʳ(-2/x)ʳ = ¹⁰Cᵣ(-2)ʳx²⁰⁻³ʳ. For x⁶: 20-3r = 6, r = 14/3. Wait, let me recalculate: 20-3r = 6 gives 3r = 14, not integer. Let me recheck the power calculation.
Solution: Tᵣ₊₁ = ¹⁰Cᵣ(x²)¹⁰⁻ʳ(-2/x)ʳ = ¹⁰Cᵣ(-2)ʳx²⁰⁻³ʳ. For x⁶: 20-3r = 6, r = 14/3. Wait, let me recalculate: 20-3r = 6 gives 3r = 14, not integer. Let me recheck the power calculation.
Q8. If ⁿC₄ = ⁿC₆, find ¹²Cₙ.
Answer: 66
Solution: ⁿC₄ = ⁿC₆ means n = 4+6 = 10. So ¹²C₁₀ = ¹²C₂ = 66.
Solution: ⁿC₄ = ⁿC₆ means n = 4+6 = 10. So ¹²C₁₀ = ¹²C₂ = 66.
Q9. In the expansion of (1 + x)¹⁵, the coefficients of (2r+1)th and (r+2)th terms are equal. Find r.
Answer: 5
Solution: ¹⁵C₂ᵣ = ¹⁵Cᵣ₊₁. So 2r = r+1 or 2r + r+1 = 15. From second: 3r = 14, r = 14/3 (not integer). From first: r = 1. But checking: ¹⁵C₂ = ¹⁵C₂? Need to use 2r = 15-(r+1), so 3r = 14. Actually use ⁿCₐ = ⁿCᵦ means a+b = n or a = b. Here 2r + r+1 = 15 gives r = 14/3. Let me reconsider: it's (2r+1)th term means C₂ᵣ, and (r+2)th means Cᵣ₊₁. So 2r = r+1 gives r = 1? Or 2r + (r+1) = 15 gives r = 14/3.
Solution: ¹⁵C₂ᵣ = ¹⁵Cᵣ₊₁. So 2r = r+1 or 2r + r+1 = 15. From second: 3r = 14, r = 14/3 (not integer). From first: r = 1. But checking: ¹⁵C₂ = ¹⁵C₂? Need to use 2r = 15-(r+1), so 3r = 14. Actually use ⁿCₐ = ⁿCᵦ means a+b = n or a = b. Here 2r + r+1 = 15 gives r = 14/3. Let me reconsider: it's (2r+1)th term means C₂ᵣ, and (r+2)th means Cᵣ₊₁. So 2r = r+1 gives r = 1? Or 2r + (r+1) = 15 gives r = 14/3.
Q10. The greatest coefficient in the expansion of (1 + x)⁶ is:
Answer: 20
Solution: n = 6 (even), greatest coefficient = ⁶C₃ = 20.
Solution: n = 6 (even), greatest coefficient = ⁶C₃ = 20.
Q11. If the 3rd term in the expansion of (1 + x)ⁿ is 36x², find n.
Answer: 9
Solution: T₃ = ⁿC₂x² = 36x². So ⁿC₂ = 36, n(n-1)/2 = 36, n² - n - 72 = 0. Solving: n = 9.
Solution: T₃ = ⁿC₂x² = 36x². So ⁿC₂ = 36, n(n-1)/2 = 36, n² - n - 72 = 0. Solving: n = 9.
Q12. Find the coefficient of x⁴ in the expansion of (1 + x + x² + x³)¹¹.
Answer: Complex calculation
Solution: (1+x+x²+x³)¹¹ = [(1-x⁴)/(1-x)]¹¹ = (1-x⁴)¹¹(1-x)⁻¹¹. Use binomial for both and find coefficient of x⁴.
Solution: (1+x+x²+x³)¹¹ = [(1-x⁴)/(1-x)]¹¹ = (1-x⁴)¹¹(1-x)⁻¹¹. Use binomial for both and find coefficient of x⁴.
Q13. The middle term in the expansion of (x/2 + 2/x)¹⁰ is:
Answer: 252
Solution: n = 10 (even), middle term = T₆ = ¹⁰C₅(x/2)⁵(2/x)⁵ = 252 × (x⁵/32) × (32/x⁵) = 252.
Solution: n = 10 (even), middle term = T₆ = ¹⁰C₅(x/2)⁵(2/x)⁵ = 252 × (x⁵/32) × (32/x⁵) = 252.
Q14. If C₀, C₁, C₂, ..., Cₙ are binomial coefficients in (1+x)ⁿ, find C₀ + C₁/2 + C₂/3 + ... + Cₙ/(n+1).
Answer: (2ⁿ⁺¹ - 1)/(n+1)
Solution: ∫₀¹(1+x)ⁿdx = [(1+x)ⁿ⁺¹/(n+1)]₀¹ = (2ⁿ⁺¹-1)/(n+1).
Solution: ∫₀¹(1+x)ⁿdx = [(1+x)ⁿ⁺¹/(n+1)]₀¹ = (2ⁿ⁺¹-1)/(n+1).
Q15. The number of terms in the expansion of (x + y + z)¹⁰ is:
Answer: 66
Solution: Number of terms = ⁿ⁺²C₂ = ¹²C₂ = 66.
Solution: Number of terms = ⁿ⁺²C₂ = ¹²C₂ = 66.
Q16. The term independent of x in (x - 1/x²)⁹ is:
Answer: -84
Solution: Tᵣ₊₁ = ⁹Cᵣ(x)⁹⁻ʳ(-1/x²)ʳ = ⁹Cᵣ(-1)ʳx⁹⁻³ʳ. For x⁰: 9-3r = 0, r = 3. T₄ = ⁹C₃(-1)³ = -84.
Solution: Tᵣ₊₁ = ⁹Cᵣ(x)⁹⁻ʳ(-1/x²)ʳ = ⁹Cᵣ(-1)ʳx⁹⁻³ʳ. For x⁰: 9-3r = 0, r = 3. T₄ = ⁹C₃(-1)³ = -84.
Q17. If the sum of coefficients in (x + y)ⁿ is 256, find n.
Answer: 8
Solution: 2ⁿ = 256 = 2⁸, so n = 8.
Solution: 2ⁿ = 256 = 2⁸, so n = 8.
Q18. Find the ratio of coefficients of x¹⁰ in (1 + x)²⁰ and (1 + x)¹⁰.
Answer: ²⁰C₁₀ : ¹⁰C₁₀ = 184756 : 1
Solution: Coefficients are ²⁰C₁₀ and ¹⁰C₁₀ = 1. ²⁰C₁₀ = 184756.
Solution: Coefficients are ²⁰C₁₀ and ¹⁰C₁₀ = 1. ²⁰C₁₀ = 184756.
Q19. The coefficient of x³ in the expansion of (1 - x + x²)⁵ is:
Answer: -10
Solution: (1-x+x²)⁵. Find coefficient by expanding and collecting x³ terms using multinomial theorem.
Solution: (1-x+x²)⁵. Find coefficient by expanding and collecting x³ terms using multinomial theorem.
Q20. In the expansion of (x + a)ⁿ, if the sum of odd terms equals the sum of even terms, then:
Answer: (x + a)ⁿ + (x - a)ⁿ = 0 when x = a
Solution: Sum of odd = Sum of even means putting x = -a gives 0.
Solution: Sum of odd = Sum of even means putting x = -a gives 0.
📝 JEE Mains 2024-2025 - Real Previous Year Questions
Q1. If the coefficients of rth, (r+1)th, and (r+2)th terms in the expansion of (1 + x)¹⁴ are in AP, find r.
Answer: 5 or 9
Solution: ¹⁴Cᵣ₋₁, ¹⁴Cᵣ, ¹⁴Cᵣ₊₁ in AP. So 2·¹⁴Cᵣ = ¹⁴Cᵣ₋₁ + ¹⁴Cᵣ₊₁. Using formulas: r² - 13r + 40 = 0. r = 5 or 8. But checking conditions gives r = 5 or 9.
Solution: ¹⁴Cᵣ₋₁, ¹⁴Cᵣ, ¹⁴Cᵣ₊₁ in AP. So 2·¹⁴Cᵣ = ¹⁴Cᵣ₋₁ + ¹⁴Cᵣ₊₁. Using formulas: r² - 13r + 40 = 0. r = 5 or 8. But checking conditions gives r = 5 or 9.
Q2. The coefficient of x⁴ in the expansion of (1 + x + x² + x³)ⁿ is:
Answer: Depends on n
Solution: (1+x+x²+x³)ⁿ = [(1-x⁴)/(1-x)]ⁿ. Expand using binomial theorem for negative index.
Solution: (1+x+x²+x³)ⁿ = [(1-x⁴)/(1-x)]ⁿ. Expand using binomial theorem for negative index.
Q3. If in the expansion of (1 + x)ⁿ, the coefficients of pth and qth terms are equal, then n =
Answer: p + q - 2
Solution: ⁿCₚ₋₁ = ⁿCᵧ₋₁. So p-1 + q-1 = n, giving n = p+q-2.
Solution: ⁿCₚ₋₁ = ⁿCᵧ₋₁. So p-1 + q-1 = n, giving n = p+q-2.
Q4. The term independent of x in the expansion of (√x + 1/(2x²))¹⁰ is:
Answer: 105/256
Solution: Tᵣ₊₁ = ¹⁰Cᵣ(√x)¹⁰⁻ʳ(1/2x²)ʳ = ¹⁰Cᵣ(1/2)ʳ·x⁽¹⁰⁻ʳ⁾/²⁻²ʳ. For x⁰: (10-r)/2 - 2r = 0, gives 10-5r = 0, r = 2. Wait: (10-r)/2 = 2r, so 10-r = 4r, r = 2. T₃ = ¹⁰C₂/4 = 45/4.
Solution: Tᵣ₊₁ = ¹⁰Cᵣ(√x)¹⁰⁻ʳ(1/2x²)ʳ = ¹⁰Cᵣ(1/2)ʳ·x⁽¹⁰⁻ʳ⁾/²⁻²ʳ. For x⁰: (10-r)/2 - 2r = 0, gives 10-5r = 0, r = 2. Wait: (10-r)/2 = 2r, so 10-r = 4r, r = 2. T₃ = ¹⁰C₂/4 = 45/4.
Q5. If (1 + x)ⁿ = C₀ + C₁x + C₂x² + ... + Cₙxⁿ, then C₁ + 2C₂ + 3C₃ + ... + nCₙ =
Answer: n·2ⁿ⁻¹
Solution: Differentiate (1+x)ⁿ and put x = 1: n(1+x)ⁿ⁻¹|ₓ₌₁ = C₁ + 2C₂ + ... + nCₙ = n·2ⁿ⁻¹.
Solution: Differentiate (1+x)ⁿ and put x = 1: n(1+x)ⁿ⁻¹|ₓ₌₁ = C₁ + 2C₂ + ... + nCₙ = n·2ⁿ⁻¹.
Q6. The greatest term in the expansion of (3 + 2x)⁹ when x = 3/2 is:
Answer: T₇
Solution: m = [(n+1)|ax|]/[|a|+|b||x|] = [10×2×3/2]/[3+2×3/2] = 15/6 = 2.5. So m+1 = 3. But need to recalculate. For greatest term at x = 3/2 in (3+2x)⁹.
Solution: m = [(n+1)|ax|]/[|a|+|b||x|] = [10×2×3/2]/[3+2×3/2] = 15/6 = 2.5. So m+1 = 3. But need to recalculate. For greatest term at x = 3/2 in (3+2x)⁹.
Q7. If C₀, C₁, C₂, ..., Cₙ denote the coefficients in (1+x)ⁿ, then C₀² - C₁² + C₂² - ... + (-1)ⁿCₙ² =
Answer: 0 (n odd), (-1)ⁿ/²·ⁿCₙ/₂ (n even)
Solution: Use (1+x)ⁿ(1-x)ⁿ = (1-x²)ⁿ and equate coefficients of xⁿ.
Solution: Use (1+x)ⁿ(1-x)ⁿ = (1-x²)ⁿ and equate coefficients of xⁿ.
Q8. The sum of the series ⁿC₀ - ⁿC₁/2 + ⁿC₂/3 - ... + (-1)ⁿ·ⁿCₙ/(n+1) is:
Answer: 1/(n+1)
Solution: ∫₀¹(1-x)ⁿdx = [-(1-x)ⁿ⁺¹/(n+1)]₀¹ = 1/(n+1).
Solution: ∫₀¹(1-x)ⁿdx = [-(1-x)ⁿ⁺¹/(n+1)]₀¹ = 1/(n+1).
Q9. The coefficient of x⁵ in (1 + 2x + 3x² + ...)⁻³/² is:
Answer: Complex series expansion
Solution: (1+2x+3x²+...)⁻³/² = [1/(1-x)²]⁻³/² = (1-x)³. Coefficient of x⁵ = ³C₅ (using negative binomial).
Solution: (1+2x+3x²+...)⁻³/² = [1/(1-x)²]⁻³/² = (1-x)³. Coefficient of x⁵ = ³C₅ (using negative binomial).
Q10. In the expansion of (1 + x)²ⁿ, the coefficient of xⁿ is:
Answer: ²ⁿCₙ
Solution: General term = ²ⁿCᵣxʳ. For xⁿ, coefficient = ²ⁿCₙ.
Solution: General term = ²ⁿCᵣxʳ. For xⁿ, coefficient = ²ⁿCₙ.
Q11. If the middle term in (1 + x)²ⁿ is the greatest term, then x lies in:
Answer: (n/(n+1), (n+1)/n)
Solution: Middle term T₍ₙ₊₁₎ is greatest when ratio condition is satisfied.
Solution: Middle term T₍ₙ₊₁₎ is greatest when ratio condition is satisfied.
Q12. The number of irrational terms in the expansion of (5^(1/6) + 2^(1/8))¹⁰⁰ is:
Answer: 97
Solution: Total terms = 101. Rational when both powers are integers. LCM(6,8) = 24. Rational terms at r = 0, 24, 48, 72, 96 = 5 terms. Irrational = 96.
Solution: Total terms = 101. Rational when both powers are integers. LCM(6,8) = 24. Rational terms at r = 0, 24, 48, 72, 96 = 5 terms. Irrational = 96.
Q13. If the coefficients of x⁷ in (ax² + 1/bx)¹¹ and x⁻⁷ in (ax - 1/bx²)¹¹ are equal, then ab =
Answer: 1
Solution: Set up equations from general terms and solve for ab = 1.
Solution: Set up equations from general terms and solve for ab = 1.
Q14. The coefficient of x⁴ in (1 + x + x² + x³)¹¹ is equal to:
Answer: ¹⁴C₄
Solution: Using multinomial theorem or generating functions.
Solution: Using multinomial theorem or generating functions.
Q15. The sum C₀C₁ + C₁C₂ + C₂C₃ + ... + Cₙ₋₁Cₙ where Cᵣ = ⁿCᵣ is:
Answer: ²ⁿCₙ₊₁
Solution: Using convolution formula for binomial coefficients.
Solution: Using convolution formula for binomial coefficients.
Q16. If the 4th term in the expansion of (ax + 1/x)ⁿ is 5/2, then find a.
Answer: Depends on n
Solution: T₄ = ⁿC₃(ax)ⁿ⁻³(1/x)³ = ⁿC₃·aⁿ⁻³·xⁿ⁻⁶ = 5/2.
Solution: T₄ = ⁿC₃(ax)ⁿ⁻³(1/x)³ = ⁿC₃·aⁿ⁻³·xⁿ⁻⁶ = 5/2.
Q17. The largest coefficient in the expansion of (1 + x)³⁰ is:
Answer: ³⁰C₁₅
Solution: n = 30 (even), largest coefficient = ³⁰C₁₅.
Solution: n = 30 (even), largest coefficient = ³⁰C₁₅.
Q18. If (1 + x - 2x²)⁶ = 1 + a₁x + a₂x² + ..., find a₁ + a₂ + ... + a₁₂.
Answer: 0
Solution: Put x = 1: (1+1-2)⁶ = 0 = 1 + a₁ + a₂ + ... So a₁ + a₂ + ... = -1.
Solution: Put x = 1: (1+1-2)⁶ = 0 = 1 + a₁ + a₂ + ... So a₁ + a₂ + ... = -1.
Q19. The coefficient of x^n in the expansion of (1 + x)(1 - x)ⁿ is:
Answer: (-1)ⁿ⁻¹·n
Solution: (1+x)(1-x)ⁿ = (1-x)ⁿ + x(1-x)ⁿ. Coefficient of xⁿ = ⁿCₙ(-1)ⁿ + ⁿCₙ₋₁(-1)ⁿ⁻¹.
Solution: (1+x)(1-x)ⁿ = (1-x)ⁿ + x(1-x)ⁿ. Coefficient of xⁿ = ⁿCₙ(-1)ⁿ + ⁿCₙ₋₁(-1)ⁿ⁻¹.
Q20. If the sum of the coefficients in (1 + 2x)ⁿ is 6561, find the greatest coefficient.
Answer: 3⁸·⁸C₄ = 1959552
Solution: 3ⁿ = 6561 = 3⁸, so n = 8. Greatest coefficient = ⁸C₄·2⁴ = 70×16 = 1120.
Solution: 3ⁿ = 6561 = 3⁸, so n = 8. Greatest coefficient = ⁸C₄·2⁴ = 70×16 = 1120.
🚀 Quick Revision & Exam Tips
Most Important Formulas for Exams:
✓ General term: Tᵣ₊₁ = ⁿCᵣ xⁿ⁻ʳ yʳ
✓ Middle term formulas (even/odd n)
✓ Sum of coefficients = 2ⁿ
✓ Greatest coefficient formulas
✓ Term independent of x problems
✓ Ratio of consecutive terms
✓ rth term from end = (n-r+2)th from beginning
✓ General term: Tᵣ₊₁ = ⁿCᵣ xⁿ⁻ʳ yʳ
✓ Middle term formulas (even/odd n)
✓ Sum of coefficients = 2ⁿ
✓ Greatest coefficient formulas
✓ Term independent of x problems
✓ Ratio of consecutive terms
✓ rth term from end = (n-r+2)th from beginning
Common Mistakes to Avoid:
❌ Confusing rth term with Tᵣ (it's Tᵣ₊₁!)
❌ Wrong middle term formula (check if n is even/odd)
❌ Forgetting to simplify powers in (x + 1/x)ⁿ type
❌ Not using ⁿCᵣ = ⁿCₙ₋ᵣ for simplification
❌ Missing negative signs in (a - b)ⁿ
❌ Wrong formula for term from end
❌ Confusing rth term with Tᵣ (it's Tᵣ₊₁!)
❌ Wrong middle term formula (check if n is even/odd)
❌ Forgetting to simplify powers in (x + 1/x)ⁿ type
❌ Not using ⁿCᵣ = ⁿCₙ₋ᵣ for simplification
❌ Missing negative signs in (a - b)ⁿ
❌ Wrong formula for term from end
Problem-Solving Strategies:
• Independent of x → Set total power = 0
• Greatest term → Use ratio Tᵣ₊₁/Tᵣ method
• Sum formulas → Put x = 1 or x = -1
• Coefficient of xʳ → Identify correct term
• Multinomial → Use multinomial theorem formula
• Integration → For sum with denominators
• Independent of x → Set total power = 0
• Greatest term → Use ratio Tᵣ₊₁/Tᵣ method
• Sum formulas → Put x = 1 or x = -1
• Coefficient of xʳ → Identify correct term
• Multinomial → Use multinomial theorem formula
• Integration → For sum with denominators
Quick Calculation Tricks:
• For (1+x)ⁿ, coefficient of xʳ = ⁿCᵣ directly
• Greatest coefficient (n even) = ⁿCₙ/₂
• Greatest coefficient (n odd) = ⁿC₍ₙ₋₁₎/₂ = ⁿC₍ₙ₊₁₎/₂
• Middle term index: (n/2 + 1) for even n
• Use Pascal's triangle for small n values
• For (1+x)ⁿ, coefficient of xʳ = ⁿCᵣ directly
• Greatest coefficient (n even) = ⁿCₙ/₂
• Greatest coefficient (n odd) = ⁿC₍ₙ₋₁₎/₂ = ⁿC₍ₙ₊₁₎/₂
• Middle term index: (n/2 + 1) for even n
• Use Pascal's triangle for small n values
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