Expand (x + y)ⁿ with the Binomial Theorem and Pascal's triangle

Learning goal: Know and apply the Binomial Theorem for the expansion of (x + y)ⁿ in powers of x and y for a positive integer n, where x and y are any numbers, using coefficients read from Pascal's triangle or computed as combinations; and understand the two ways the theorem can be proved, by a combinatorial argument or by mathematical induction.

Scope note. A.APR.5 is marked (+) in Ohio's Learning Standards for mathematics, which Ohio uses to indicate a standard "for students who are planning on taking advanced courses". It is taught here in full, as extension content for students on that path.

Expansions of small powers. Multiplying out by hand gives (x + y)¹ = x + y; (x + y)² = x² + 2xy + y²; (x + y)³ = x³ + 3x²y + 3xy² + y³; and (x + y)⁴ = x⁴ + 4x³y + 6x²y² + 4xy³ + y⁴. Three patterns appear. Every term has total degree n: the exponent of x starts at n and falls by one each term while the exponent of y starts at 0 and rises by one, so the exponents in each term add to n. There are n + 1 terms. And the coefficients, 1 2 1, then 1 3 3 1, then 1 4 6 4 1, are the rows of Pascal's triangle.

Pascal's triangle. The triangle begins with a single 1 in row 0. Each later row starts and ends with 1, and every interior entry is the sum of the two entries diagonally above it. Row 1 is 1 1; row 2 is 1 2 1; row 3 is 1 3 3 1; row 4 is 1 4 6 4 1; row 5 is 1 5 10 10 5 1, because 1 + 4 = 5, 4 + 6 = 10, 6 + 4 = 10 and 4 + 1 = 5. The diagram shows rows 0 to 5. Row n gives the coefficients of (x + y)ⁿ.

Combinations. The entry in row n, position k (counting from k = 0) is the number of ways to choose k objects from n, written C(n, k) and equal to n! / (k!(n − k)!), where n! means n × (n − 1) × … × 1 and 0! = 1. For example C(5, 2) = 120 / (2 × 6) = 10, which is the third entry of row 5. The Pascal rule C(n, k − 1) + C(n, k) = C(n + 1, k) is exactly the "add the two above" rule, and it is why the triangle and the formula agree.

The Binomial Theorem. For any numbers x and y and any positive integer n, (x + y)ⁿ = C(n, 0)xⁿ + C(n, 1)xⁿ⁻¹y + C(n, 2)xⁿ⁻²y² + … + C(n, n − 1)xyⁿ⁻¹ + C(n, n)yⁿ. In words: the coefficient of xⁿ⁻ᵏyᵏ is C(n, k). The condition "positive integer n" matters; for fractional or negative exponents the expansion has infinitely many terms and is a different theorem.

Why it is true: the combinatorial argument. Write (x + y)ⁿ as a product of n identical factors (x + y)(x + y)…(x + y). When it is expanded without collecting, each term is formed by choosing x or y from every factor and multiplying the choices. A term equals xⁿ⁻ᵏyᵏ exactly when y was chosen from k of the n factors and x from the rest. The number of ways to choose which k factors supply the y is C(n, k). So after collecting, xⁿ⁻ᵏyᵏ has coefficient C(n, k). That is the whole proof.

Why it is true: the inductive argument in outline. The theorem holds for n = 1. Suppose it holds for some n. Then (x + y)ⁿ⁺¹ = (x + y)(x + y)ⁿ; multiplying the assumed expansion by x shifts every coefficient one place toward higher powers of x, and multiplying by y shifts toward higher powers of y, and adding the two shifted rows produces the entries C(n, k − 1) + C(n, k) = C(n + 1, k). That is Pascal's rule, so the theorem holds for n + 1. Because it holds for 1 and each case implies the next, it holds for every positive integer n.

A fully worked model with substitution. Expand (2a − 3)³. This is (x + y)³ with x = 2a and y = −3. The row-3 coefficients are 1 3 3 1, so (x + y)³ = x³ + 3x²y + 3xy² + y³. Substitute: (2a)³ = 8a³; 3(2a)²(−3) = 3·4a²·(−3) = −36a²; 3(2a)(−3)² = 3·2a·9 = 54a; (−3)³ = −27. So (2a − 3)³ = 8a³ − 36a² + 54a − 27. The step students skip is raising the whole of 2a to the power, coefficient included, and keeping the sign of −3 inside its power. Check at a = 2: (4 − 3)³ = 1, and 64 − 144 + 108 − 27 = 1.

Numerical uses. Because x and y may be any numbers, the theorem computes powers of sums. 11⁴ = (10 + 1)⁴ = 10⁴ + 4·10³ + 6·10² + 4·10 + 1 = 10000 + 4000 + 600 + 40 + 1 = 14641, which is row 4 of the triangle read as a number. And (1.01)⁵ = (1 + 0.01)⁵ = 1 + 5(0.01) + 10(0.0001) + 10(0.000001) + 5(0.00000001) + 0.0000000001 = 1.0510100501, where the first two or three terms already give an excellent approximation. Setting x = y = 1 gives 2ⁿ as the sum of row n, so row 5 sums to 32; setting x = 1, y = −1 gives 0, so the alternating sum of every row from row 1 on is 0. Row 0 is different: it is the single coefficient C(0, 0) = 1, the one way to choose nothing from nothing, so its alternating sum is 1. (The substitution argument uses n ≥ 1, the positive integer exponents of the theorem.)

Finding one term without the whole expansion. The coefficient of x³y⁴ in (x + y)⁷ is C(7, 4) = 35, and it is also C(7, 3), since choosing which 4 factors give y is the same as choosing which 3 give x. For (x − y)⁷, the x⁵y² term is C(7, 2)x⁵(−y)² = 21x⁵y², positive because the even power of −y is positive; the x⁴y³ term would be C(7, 3)x⁴(−y)³ = −35x⁴y³.

The misconception to remove. Students write (x + y)ⁿ = xⁿ + yⁿ. For n = 2 and x = y = 1 that claims 4 = 2. The middle terms are real, and Pascal's triangle counts exactly how many of each there are.

Worked full expansion: Expand (x + 2)⁵. Row 5 of Pascal's triangle is 1 5 10 10 5 1, and y = 2 with powers 1, 2, 4, 8, 16, 32. Terms: 1·x⁵·1 = x⁵; 5·x⁴·2 = 10x⁴; 10·x³·4 = 40x³; 10·x²·8 = 80x²; 5·x·16 = 80x; 1·32 = 32. So (x + 2)⁵ = x⁵ + 10x⁴ + 40x³ + 80x² + 80x + 32. Check at x = 1: 3⁵ = 243, and 1 + 10 + 40 + 80 + 80 + 32 = 243.

Worked expansion with a coefficient inside: Expand (1 + 2x)⁴. Row 4 is 1 4 6 4 1, with x = 1 and y = 2x. Terms: 1; 4(2x) = 8x; 6(2x)² = 24x²; 4(2x)³ = 32x³; (2x)⁴ = 16x⁴. Result: 1 + 8x + 24x² + 32x³ + 16x⁴. Check at x = 1: 3⁴ = 81 and 1 + 8 + 24 + 32 + 16 = 81. Note that the coefficient 2 is raised to the power along with x.

Worked single term: Find the x² term of (x − 1)⁶. The general term is C(6, k)x⁶⁻ᵏ(−1)ᵏ, and x² needs 6 − k = 2, so k = 4. C(6, 4) = 720 / (24 × 2) = 15, and (−1)⁴ = 1, so the term is 15x². Row 6 of the triangle, 1 6 15 20 15 6 1, confirms the 15 in position 4.

Worked numerical application: Compute 1.02³ using (1 + 0.02)³ = 1 + 3(0.02) + 3(0.02)² + (0.02)³ = 1 + 0.06 + 0.0012 + 0.000008 = 1.061208. Direct multiplication, 1.02 × 1.02 = 1.0404 and 1.0404 × 1.02 = 1.061208, agrees. The theorem turns a power of a sum into a short list of terms that shrink quickly, which is why it is used for approximations.

Representation task: Write Pascal's triangle from row 0 to row 7, showing with small arrows how three interior entries of row 6 were obtained from row 5. Beside rows 2, 3 and 4 write the corresponding expansions of (x + y)ⁿ with every term. Then compute C(6, 2), C(6, 3) and C(7, 3) from the formula n! / (k!(n − k)!) and mark where each sits in your triangle, showing that the formula and the addition rule agree.

Independent performance: Expand (3x − 2)⁴ and (a + b²)⁵ completely using the Binomial Theorem, then find, without full expansion, the coefficient of x⁴y³ in (x + y)⁷ and the x³ term in (2x − 1)⁶. For one of the full expansions, verify by substituting a value. Then write a paragraph giving the combinatorial reason the coefficient of xⁿ⁻ᵏyᵏ is C(n, k). Success criteria: every term carries the correct power of the inside coefficient and the correct sign; the substitution check matches; the single-term answers name n and k explicitly; and the paragraph refers to choosing which factors contribute y.