Binomial expansion is a fundamental tool in architectural mathematics, allowing precise calculation of expressions raised to powers, which frequently appear in structural load analysis, material strength estimations, and geometric design computations. Mastery of binomial expansion aids architects in simplifying complex polynomial expressions that model real-world architectural elements, such as curved surfaces or load distributions. This chapter focuses on the mathematical techniques behind binomial expansions, starting with Pascal’s triangle and advancing to expansions involving fractional and negative powers, crucial for advanced architectural calculations.

2.1 Binomial Theorem

The binomial theorem is a formula for expanding any power of a binomial, an expression with two terms such as a + b, without multiplying the brackets out one by one. Multiplying (a + b) by itself four or five times by hand is slow and easy to get wrong, while the theorem gives every term directly from a simple pattern of powers and coefficients. It is used in algebra, probability, statistics, finance and engineering wherever powers of two-term expressions appear.

2.1.1 Expansion of (a + b)ⁿ using the binomial theorem

The binomial theorem states that, for any positive whole number n, the expansion of (a + b)ⁿ is the sum of n + 1 terms in which the power of a falls from n to 0, the power of b rises from 0 to n, and each term is multiplied by a binomial coefficient. Written out in full:

(a + b)ⁿ = nC0 aⁿ + nC1 aⁿ⁻¹ b + nC2 aⁿ⁻² b² + nC3 aⁿ⁻³ b³ + ... + nCn bⁿ

The general term, which gives any single term without writing the whole expansion, is:

T(r + 1) = nCr × aⁿ⁻ʳ × bʳ, for r = 0, 1, 2, ..., n

Example: a quantity surveyor checking a formula involving (1 + x)⁵ can write all six terms in one line using the theorem instead of multiplying five brackets together.

Features of a binomial expansion

• Number of terms: The expansion of (a + b)ⁿ has n + 1 terms, so (a + b)⁴ has 5 terms.
• Powers of a: The power of a starts at n in the first term and falls by 1 in each term until it is 0.
• Powers of b: The power of b starts at 0 and rises by 1 in each term until it reaches n.
• Sum of powers: In every term, the power of a plus the power of b equals n.
• Symmetry of coefficients: The coefficients read the same from the start and from the end, for example 1, 4, 6, 4, 1.
• Negative second term: For (a − b)ⁿ, write it as (a + (−b))ⁿ, so the signs alternate +, −, +, −.

Steps for expanding (a + b)ⁿ

1. Identify a, b and n: For (2x − 3)³, a = 2x, b = −3 and n = 3, keeping the sign with b.
2. Count the terms: There will be n + 1 terms in the expansion.
3. Write the coefficients: Calculate nC0, nC1, ..., nCn for each term.
4. Write the powers of a: Start with aⁿ and reduce the power by 1 in each term.
5. Write the powers of b: Start with b⁰ = 1 and increase the power by 1 in each term.
6. Multiply each term out: Multiply the coefficient, the power of a and the power of b, taking care with signs and numbers inside brackets.
7. Simplify and check: Collect the terms and check that the powers in each term add up to n.

Worked example 1: expanding (x + 2)⁴

a = x, b = 2, n = 4, coefficients 1, 4, 6, 4, 1

(x + 2)⁴ = 1(x⁴) + 4(x³)(2) + 6(x²)(2²) + 4(x)(2³) + 1(2⁴)

= x⁴ + 8x³ + 6(x²)(4) + 4(x)(8) + 16

= x⁴ + 8x³ + 24x² + 32x + 16

Worked example 2: expanding (2x − 3)³

a = 2x, b = −3, n = 3, coefficients 1, 3, 3, 1

(2x − 3)³ = 1(2x)³ + 3(2x)²(−3) + 3(2x)(−3)² + 1(−3)³

= 8x³ + 3(4x²)(−3) + 3(2x)(9) + (−27)

= 8x³ − 36x² + 54x − 27

Worked example 3: expanding (1 + x)⁵

a = 1, b = x, n = 5, coefficients 1, 5, 10, 10, 5, 1

(1 + x)⁵ = 1 + 5x + 10x² + 10x³ + 5x⁴ + x⁵

Worked example 4: finding a particular term

Find the 4th term of (2x + 1)⁶.

The 4th term is T(r + 1) with r = 3.

T(4) = 6C3 × (2x)⁶⁻³ × 1³

T(4) = 20 × (2x)³ × 1

T(4) = 20 × 8x³

T(4) = 160x³

Worked example 5: finding a coefficient

Find the coefficient of x³ in (3 + 2x)⁵.

The term containing x³ has r = 3.

T(4) = 5C3 × 3⁵⁻³ × (2x)³

T(4) = 10 × 3² × 8x³

T(4) = 10 × 9 × 8x³

T(4) = 720x³

The coefficient of x³ is 720.

2.1.2 Binomial coefficients and their calculation

Binomial coefficients are the numbers that multiply each term in a binomial expansion, written nCr (read "n choose r") or as n over r in brackets. Each one counts the number of ways of choosing r items from n, which is why the same numbers appear in expansions and in counting problems. They are calculated with factorials, where n! (n factorial) means n × (n − 1) × (n − 2) × ... × 2 × 1, and 0! is defined as 1. Example: a SACCO with 7 board members wishing to form a 3-member audit committee can do so in 7C3 = 35 different ways.

The formula for a binomial coefficient is:

nCr = n! ÷ (r! × (n − r)!)

Properties of binomial coefficients

• End coefficients: nC0 = 1 and nCn = 1, so every expansion starts and ends with a coefficient of 1.
• Second coefficient: nC1 = n, so the second term's coefficient equals the power.
• Symmetry: nCr = nC(n − r), for example 7C2 = 7C5 = 21.
• Sum of coefficients: nC0 + nC1 + ... + nCn = 2ⁿ, found by putting a = b = 1.
• Whole numbers: Binomial coefficients are always positive whole numbers.
• Calculator key: Scientific calculators have an nCr key: enter n, press nCr, enter r, press =.

Steps for calculating nCr by formula

1. Write the formula: nCr = n! ÷ (r! × (n − r)!).
2. Substitute n and r: Replace n and r with the given numbers.
3. Find n − r: Work out the value of n − r.
4. Cancel the larger factorial: Write n! as a product down to the larger of r! or (n − r)! and cancel.
5. Multiply out: Multiply the remaining numbers on the top and on the bottom.
6. Divide: Divide the top by the bottom to get a whole number.
7. Check: Use the symmetry property or a calculator to confirm the answer.

Worked example 1: 5C2

5C2 = 5! ÷ (2! × 3!)

5C2 = (5 × 4 × 3!) ÷ (2 × 1 × 3!)

5C2 = (5 × 4) ÷ 2

5C2 = 20 ÷ 2

5C2 = 10

Worked example 2: 7C3

7C3 = 7! ÷ (3! × 4!)

7C3 = (7 × 6 × 5 × 4!) ÷ (3 × 2 × 1 × 4!)

7C3 = 210 ÷ 6

7C3 = 35

Worked example 3: 10C4

10C4 = 10! ÷ (4! × 6!)

10C4 = (10 × 9 × 8 × 7) ÷ (4 × 3 × 2 × 1)

10C4 = 5 040 ÷ 24

10C4 = 210

Worked example 4: using symmetry for 9C7

9C7 = 9C(9 − 7) = 9C2

9C2 = (9 × 8) ÷ (2 × 1)

9C2 = 72 ÷ 2

9C2 = 36

Worked example 5: checking the sum of coefficients

For n = 5, the coefficients are 5C0, 5C1, 5C2, 5C3, 5C4 and 5C5.

5C0 = 1, 5C1 = 5, 5C2 = 10, 5C3 = 10, 5C4 = 5, 5C5 = 1

Sum = 1 + 5 + 10 + 10 + 5 + 1

Sum = 32

2⁵ = 32, which confirms the property.

Chapter Summary

This chapter focused on carrying out binomial expansion by first introducing the binomial theorem for expanding expressions of the form \((a + b)^n\). It explained the calculation of binomial coefficients, which play a key role in the expansion process. The relationship between Pascal's Triangle and binomial coefficients was explored, showing how the triangle provides a simple method to determine these coefficients. The chapter also covered the expansion of binomial expressions with fractional and negative powers, extending the theorem beyond positive integers. Errors that arise in binomial expansion were discussed, including how the expansion can be used to approximate values and estimate small changes effectively. Finally, practical applications were emphasized, particularly in financial calculations such as compound interest and in minimizing errors in physics and engineering problems, demonstrating the theorem’s real-life significance.