68 free MCQs on Binomial Theorem, each with its own worked answer and explanation. Expansion of (a+b) n , general term, middle term, binomial coefficients, and greatest term
Take the timed Binomial Theorem chapterwise test →68 practice questions on Binomial Theorem, sorted Easy → Hard. Try each one first, then open its answer page for the worked explanation. Want the full theory first? Read the Binomial Theorem notes.
Row n of Pascal's triangle gives the coefficients nC0, nC1, ..., nCn for the expansion of (a+b)ⁿ; each number is the sum of the two numbers diagonally above it, a direct visual proof of the identity nCr = (n-1)C(r-1) + (n-1)Cr.
Q1.
The expansion of (a + b)² is:
Q2.
The general term in the expansion of (a+b)ⁿ is:
Q5.
(1 + x)³ expanded is:
Q12.
The binomial coefficient nCr satisfies nCr = nC___.
Q13.
The second term T₂ in the expansion of (a+b)ⁿ is:
Q14.
The sum nC<sub>0</sub> − nC<sub>1</sub> + nC<sub>2</sub> − ... = ?
Q16.
Pascal triangle entry: nCr = (n-1)C(r-1) + ___.
Q17.
(a - b)⁴ = a⁴ - 4a³b + 6a²b² ___.
Q18.
Which term in (x + a)ⁿ contains b = a<sup>n</sup>?
Q21.
Find the middle term in the expansion of (x + 1/x)⁸.
Q24.
In (1+x)ⁿ, if the coefficient of x² equals the coefficient of x³, find n.
Q26.
The coefficient of x⁴ in the expansion of (1 + x + x²)⁵ is:
Q27.
The 4th term in the expansion of (x - 1/x)⁷ is:
Q29.
The greatest binomial coefficient in (1+x)¹⁰ is:
Q30.
The number of terms in (a + b + c)ⁿ is:
Q31.
Find the term containing x³ in (3 + x/2)⁸.
Q32.
If the coefficients of 2nd and 3rd terms in (1+x)ⁿ are in ratio 1:2, find n.
Q33.
The sum nC<sub>0</sub> + nC<sub>2</sub> + nC<sub>4</sub> + ... (even-indexed coefficients) =
Q35.
In (1 + x)⁵⁰, what is the ratio of the coefficient of x² to that of x⁴⁸?
Q36.
The 3rd term in the expansion of (1 - x)⁻¹ for |x| < 1 is:
Q37.
If (1 + x)ⁿ = C₀ + C₁x + C₂x² + ..., then C₀ + 2C₁ + 3C₂ + ... + (n+1)Cₙ =
Q39.
Expand (1.01)⁵ approximately using binomial theorem (first two terms):
Q40.
In the expansion of (1 + x)²⁰, the coefficient of xʳ equals the coefficient of x<sup>20-r</sup>. This is because:
Q41.
Find the term independent of x in (x² + 1/x³)¹⁰.
Q42.
If the coefficients of x⁷ and x⁸ in (2 + x/3)ⁿ are equal, find n.
Q43.
The coefficient of x⁴ in the expansion of (1 + x)ⁿ(1 + x)ⁿ = (1+x)²ⁿ is:
Q44.
The greatest coefficient in the expansion of (1 + x)²ⁿ⁺¹ is:
Q45.
nC<sub>1</sub> + 2·nC<sub>2</sub> + 3·nC<sub>3</sub> + ... + n·nCn =
Q46.
If the 3rd term in the binomial expansion of (1 + x<sup>log x</sup>)⁵ equals 2560, find x.
Q47.
Using binomial theorem, find the remainder when 3¹⁰⁰ is divided by 4.
Q48.
The sum C₀² + C₁² + C₂² + ... + Cₙ² (where Cᵣ = nCr) equals:
Q49.
In the expansion of (1 + x)ⁿ, if the coefficients of rth and (r+2)th terms are equal, then n =
Q50.
Find the value of C₀ + C₁/2 + C₂/3 + ... + Cₙ/(n+1) where Cᵣ = nCr.
Q51.
The largest term in the expansion of (3 + 5x)¹⁰ when x = 1/5 is:
Q52.
The value of (nC<sub>1</sub>/nC<sub>0</sub>) + 2(nC<sub>2</sub>/nC<sub>1</sub>) + 3(nC<sub>3</sub>/nC<sub>2</sub>) + ... + n(nCn/nCn-1) is:
Q53.
In the expansion of (1 + x)<sup>n + 2</sup>, the coefficient of x<sup>n</sup> is (n+2)C n. Using this, find the coefficient of x² in (1 + x)⁵ + (1 + x)⁶ + ... + (1 + x)¹⁰.
Q54.
Using binomial theorem, the integral part of (5 + 2√6)ⁿ + (5 - 2√6)ⁿ is:
Q55.
If the middle term in (x + 1/x)<sup>2n</sup> is 184756 nCn, find n.
Q57.
If (1 + ax)ⁿ = 1 + 6x + 12x² + ..., find a and n.
Q58.
The sum of the series 1 + nC<sub>1</sub> + (nC<sub>2</sub>)² + ... + (nCn)² is:
Q59.
Find the numerically greatest term in (2 - 3x)¹⁰ when x = 1.
Q60.
The value of C₀ - C₁ + C₂ - ... + (−1)ⁿCₙ, where Cᵣ = nCr, is:
Q61.
The coefficient of x⁵ in the expansion of (1 + x)¹⁰ is:
Q62.
The term independent of x in the expansion of (x² − 1/x)⁶ is:
Q63.
The middle term in the expansion of (2x − 1/x)⁸ is:
Q64.
The sum of all the binomial coefficients in the expansion of (2x + 3y)¹⁰ is:
Q65.
In the expansion of (1 + x)¹⁵, the coefficients of which two consecutive terms are equal?
Q66.
The value of C(10,0) + C(10,2) + C(10,4) + ... + C(10,10) is:
Q68.
If (1 + x)ⁿ = C₀ + C₁x + ... + Cₙxⁿ, then C₀ + 2C₁ + 3C₂ + ... + (n+1)Cₙ equals:
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