Questions: Binomial Theorem Expansion

Question 1

What is the coefficient of x^3 in the expansion of (x + 2)^5?

• A. 10
• B. 40
• C. 80
• D. 20

Answer: (B)

The term with x^3 requires n - k = 3, so k = 2. The term is C(5,2)·x^3·2^2 = 10·4 = 40. Option A (10) is the error of using only C(5,2) and forgetting to raise the coefficient 2 to the power k. Option C (80) comes from incorrectly setting k = 3 — as if 'x^3 means k = 3' — which actually gives the x^2 term. Always solve for k from n - k = desired exponent.

Question 2

What is the coefficient of y^3 in the expansion of (1 − y)^4?

• A. 4
• B. −4
• C. 6
• D. −6

Answer: (B)

Here a = 1, b = −y, n = 4, and k = 3. The term is C(4,3)·1^1·(−y)^3 = 4·(−1)^3·y^3 = −4y^3. The coefficient is −4. Option A (4) is the classic sign error — forgetting that (−y)^3 = −y^3. Option C and D involve using k = 2 by mistake. In any expansion of (a − b)^n, you must treat the full (−b) as the second term, so (−b)^k carries a sign of (−1)^k.

Question 3

In the expansion of (a − b)^n, the terms alternate in sign because (−b)^k is negative for odd k and positive for even k.

• T. True
• F. False

Answer: True

This is exactly correct. When you substitute −b for the second term, each term becomes C(n,k)·a^(n−k)·(−b)^k = C(n,k)·(−1)^k·a^(n−k)·b^k. Since (−1)^k = +1 for even k and −1 for odd k, the signs alternate: the k=0 term is positive, k=1 negative, k=2 positive, and so on.

Question 4

The binomial coefficient C(n,k) equals the permutation P(n,k) = n!/(n−k)!, since both count the number of ways to choose k items from n.

• T. True
• F. False

Answer: False

P(n,k) counts ordered selections; C(n,k) = n!/(k!(n−k)!) counts unordered selections. C(n,k) = P(n,k) / k! because each unordered group of k items can be arranged in k! different orders. In the binomial expansion, we want to count how many factors we pick b from — the order of the factors doesn't matter — so combinations C(n,k) are correct, not permutations.

Question 5

Explain, in terms of multiplying out the n factors of (a + b)^n, why the coefficient of a^(n−k)·b^k is C(n,k).

Model answer: When you distribute (a + b)^n, you choose one term — either a or b — from each of the n factors. To produce the monomial a^(n−k)·b^k, you must choose b from exactly k of the n factors (and a from the rest). The number of ways to choose which k factors contribute a b is C(n,k) = n!/(k!(n−k)!). Every such selection yields the same product, so C(n,k) identical terms combine into the coefficient C(n,k)·a^(n−k)·b^k.

This combinatorial argument is the conceptual heart of the theorem — the algebra is just collecting these choices. It also explains why C(n,k) appears in Pascal's triangle: the recursive identity C(n,k) = C(n−1,k−1) + C(n−1,k) says either the last factor contributes a b (C(n−1,k−1) ways to fill the rest) or it contributes an a (C(n−1,k) ways to fill the rest).