Questions: Binomial Theorem

Question 1

What is the 4th term (k = 3) in the expansion of (x - 2)^6?

• A. C(6,3) · x³ · 2³ = 160x³
• B. C(6,3) · x³ · (-2)³ = -160x³
• C. C(6,3) · x³ · 2³ = 20x³
• D. C(6,4) · x⁴ · (-2)² = 60x⁴

Answer: (B)

The (k+1)-th term is C(n,k)·a^(n-k)·b^k. Here n=6, k=3, a=x, b=-2. So: C(6,3)·x^(6-3)·(-2)^3 = 20·x³·(-8) = -160x³. The two most common errors are: (1) forgetting that b = -2, not +2, so (-2)^3 = -8, not +8; and (2) computing the wrong binomial coefficient. C(6,3) = 6!/(3!·3!) = 20, not 160. The negative sign is essential — dropping it is the typical mistake.

Question 2

How many terms are in the expansion of (a + b)^8?

• A. 8 terms — one for each power of b from b¹ to b⁸
• B. 8 terms — matching the exponent
• C. 9 terms — one for each value of k from 0 to 8
• D. 16 terms — each factor of (a + b) contributes two terms

Answer: (C)

The expansion sums over k = 0, 1, 2, ..., n — that is n+1 values, giving n+1 terms. For n=8, there are 9 terms. The k=0 term is a⁸ (no b), and the k=8 term is b⁸ (no a). A very common error is to say 'n terms' because the exponent is n, but the count is always one more than the exponent. Check with (a+b)²: it produces 3 terms (a², 2ab, b²), not 2.

Question 3

The expansion of (a + b)^n has exactly n terms.

• T. True
• F. False

Answer: False

The expansion of (a + b)^n has n+1 terms, corresponding to k = 0, 1, 2, ..., n. The off-by-one error is pervasive: students see the exponent n and expect n terms, forgetting that the k=0 term (which is aⁿ, with b absent) is still a term in the expansion. For example, (a+b)³ = a³ + 3a²b + 3ab² + b³ has 4 terms, not 3.

Question 4

In the expansion of (a + b)^n, the exponents of a and b in any single term always sum to n.

• T. True
• F. False

Answer: True

Every term has the form C(n,k)·a^(n-k)·b^k. The exponent of a is (n-k) and the exponent of b is k, so their sum is (n-k)+k = n. This reflects the combinatorial structure: each term arises from choosing a or b from each of the n factors of (a+b), and the total count of choices is always n. Every term therefore has total degree n — the expansion is homogeneous of degree n.

Question 5

Explain why the coefficient of a^(n-k)b^k in the expansion of (a + b)^n is C(n, k). What combinatorial reasoning justifies this?

Model answer: When (a+b)^n is expanded by multiplying out n copies of (a+b), each term is formed by picking either a or b from each factor. To get the term a^(n-k)b^k, you must choose b from exactly k of the n factors (and a from the remaining n-k). The number of ways to choose which k factors contribute the b is C(n,k) = n!/(k!(n-k)!). Since each such selection produces the same monomial a^(n-k)b^k, that term appears C(n,k) times in the product — giving it coefficient C(n,k).

The combinatorial interpretation is the heart of the theorem. C(n,k) isn't an arbitrary formula — it counts selections, and each selection corresponds to a distinct way of obtaining the same monomial when multiplying n factors. This also explains the symmetry C(n,k) = C(n,n-k): choosing k positions for b is equivalent to choosing n-k positions for a. Pascal's Triangle organizes these counts visually, and the identity C(n,k) = C(n-1,k-1) + C(n-1,k) mirrors the recursive structure of multiplication.