Binomial Theorem — Class 11 MCQs with Answers

Binomial Theorem — important questions & MCQs with answers (Class 11 Mathematics)

6 solved questions from this chapter's difficulty levels, each with its answer and explanation. Tap an option to check yourself.

0 of 6 answered

1. Q1 of 6Easy

   (a+b)² =

   Aa²+2ab+b²Ba²+b²Ca²−b²Da²−2ab+b²
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   Answer: a²+2ab+b²

   Explanation: The (a+b)² identity expands to a²+2ab+b² — dropping the cross term 2ab gives the tempting but wrong a²+b², which only holds when ab=0.

2. Q2 of 6Easy

   Binomial theorem: (a + b)ⁿ =

   Aaⁿ + bⁿBΣ aⁿ · bⁿCΣ C(n, r) aⁿ⁻ʳ bʳD(ab)ⁿ
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   Answer: Σ C(n, r) aⁿ⁻ʳ bʳ

   Explanation: Binomial theorem: (a+b)ⁿ = Σᵣ₌₀ⁿ ⁿCᵣ·aⁿ⁻ʳ·bʳ — dropping the ⁿCᵣ weight and just summing aⁿ·bⁿ ignores that each term has its own power split.

3. Q3 of 6Medium

   Pascal's triangle row for n = 4 gives:

   A1, 5, 10, 10, 5, 1B1, 3, 3, 1C1, 4, 6, 4, 1D1, 4, 4, 1
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   Answer: 1, 4, 6, 4, 1

   Explanation: Pascal's row for n=4 is 1,4,6,4,1, giving (a+b)⁴'s coefficients — confusing it with row n=5 (1,5,10,10,5,1) by miscounting rows is the trap.

4. Q4 of 6Medium

   Assertion (A): ⁵C₀ + ⁵C₁ + ⁵C₂ + ⁵C₃ + ⁵C₄ + ⁵C₅ = 32. Reason (R): For every non-negative integer n, ⁿC₀ = 1 = ⁿCₙ.

   ABoth A and R are true and R is the correct explanation of ABBoth A and R are true but R is not the correct explanation of ACA is true but R is falseDA is false but R is true
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   Answer: Both A and R are true but R is not the correct explanation of A

   Explanation: A is true: 1 + 5 + 10 + 10 + 5 + 1 = 32, which is 2⁵ (put x = 1 in (1 + x)⁵). R is also true, but it only gives the two end values, not the total, so R does not explain A.

5. Q5 of 6Hard

   In expansion of (x + a)ⁿ, ratio of T_r to T_r₊₁ depends on r in pattern:

   A(r+1)·x/((n−r)·a)BAlways 1Cn/rDx/a
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   Answer: (r+1)·x/((n−r)·a)

   Explanation: With Tᵣ=ⁿCᵣ·xⁿ⁻ʳ·aʳ, the ratio Tᵣ/Tᵣ₊₁ = [ⁿCᵣ/ⁿCᵣ₊₁]·(x/a) = (r+1)x/((n−r)a) — inverting this fraction to (n−r)x/((r+1)a) is the common algebra slip.

6. Q6 of 6Hard

   Assertion (A): For every positive integer n, 6ⁿ – 5n leaves remainder 1 when divided by 25. Reason (R): For every positive integer n, ⁿC₀ + ⁿC₁ + ⁿC₂ + … + ⁿCₙ = 2ⁿ.

   ABoth A and R are true and R is the correct explanation of ABBoth A and R are true but R is not the correct explanation of ACA is true but R is falseDA is false but R is true
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   Answer: Both A and R are true but R is not the correct explanation of A

   Explanation: A is true: 6ⁿ = (1 + 5)ⁿ = 1 + 5n + 25(ⁿC₂ + 5·ⁿC₃ + …), so 6ⁿ – 5n = 25k + 1. R is true (put x = 1 in (1 + x)ⁿ), but A is proved by expanding (1 + 5)ⁿ, not from the sum of the coefficients, so R does not explain A.

Binomial Theorem notes: key concepts (Class 11 Mathematics)

The binomial theorem expands (a + b)ⁿ for a positive integer n into n + 1 terms built from the coefficients nCr. This chapter derives the expansion, isolates the general term T(r+1) to reach any single term, locates the middle term, and explores the properties and sums of the binomial coefficients.

Binomial theorem

(a + b)ⁿ = Σ nCr aⁿ⁻ʳ bʳ for r = 0 to n; the powers of a fall from n to 0 while the powers of b rise from 0 to n.

Binomial coefficients

The numbers nC0, nC1, …, nCn multiplying each term; they form the nth row of Pascal's triangle and are symmetric about the centre.

Number of terms

The expansion of (a + b)ⁿ has exactly n + 1 terms, and in every term the exponents of a and b add up to n.

General term

T(r+1) = nCr aⁿ⁻ʳ bʳ is the (r+1)th term; setting the power of the variable to a target value isolates any specific term.

Middle term

If n is even there is one middle term, the (n/2 + 1)th; if n is odd there are two, the ((n+1)/2)th and ((n+3)/2)th terms.

Term independent of x

Obtained from the general term by equating the net power of x to zero, solving for r, and substituting r back to get the constant.

Sum of coefficients

Putting a = b = 1 gives nC0 + nC1 + … + nCn = 2ⁿ, the total of every binomial coefficient in the expansion.

Alternating coefficient sum

Putting a = 1, b = −1 gives nC0 − nC1 + nC2 − … = 0, so even- and odd-position coefficients each sum to 2ⁿ⁻¹.

Symmetry of coefficients

Since nCr = nC(n−r), terms equidistant from the two ends of the expansion carry identical binomial coefficients.

Term from the end

The (r+1)th term from the end of (a + b)ⁿ equals the (r+1)th term from the start of (b + a)ⁿ, useful for 'from the end' questions.

Pascal's triangle

Each coefficient is the sum of the two directly above it, since nCr = (n−1)C(r−1) + (n−1)Cr; the rows list the coefficients.

Binomial approximation

For small x, (1 + x)ⁿ ≈ 1 + nx keeps only the first two terms, giving quick estimates when higher powers are negligible.