Welding professionals frequently encounter situations requiring precise calculations to estimate dimensions, quantities, and costs. The binomial expansion is a powerful mathematical tool that simplifies the expansion of expressions raised to powers, enabling welders and fabricators to perform accurate measurements and predictions efficiently. Mastery of the binomial series and theorem supports effective problem-solving in welding design, structural calculations, and material estimation.

3.1 Binomial series

The binomial series gives a way to express powers of a binomial, an expression with two terms such as (a + b), as a sum of simpler terms. For whole-number powers the sum is finite and exact, and when one term is small it also allows quick approximations. Welding engineers use these expansions to estimate changes in areas, volumes and dimensions in structural analysis and material deformation calculations where the exact forms are tedious. Negative and fractional powers are dealt with in the binomial theorem (3.2).

3.1.1 Powers

A binomial is an algebraic expression with exactly two terms, such as (a + b), (1 + x) or (2x − 3). A power of a binomial, written (a + b)^n, means the binomial multiplied by itself n times, where n is a positive whole number. Expanding the power by repeated multiplication quickly becomes long, so we first study the pattern that every expansion follows. For example, (a + b)² = (a + b)(a + b) = a² + 2ab + b², and (a + b)³ = (a + b)(a² + 2ab + b²) = a³ + 3a²b + 3ab² + b³.

The pattern of the terms in the expansion of (a + b)^n is:

• Number of terms: the expansion has n + 1 terms. For (a + b)⁵ there are 6 terms.
• Powers of the first term: the power of a starts at n in the first term and falls by 1 in each following term until it reaches 0 in the last term.
• Powers of the second term: the power of b starts at 0 in the first term and rises by 1 in each following term until it reaches n in the last term.
• Degree of each term: in every term the power of a plus the power of b equals n. In a³b² the powers 3 + 2 = 5, so it belongs to (a + b)⁵.
• Signs: if both terms are positive, every term is positive. If the second term is negative, as in (a − b)^n, the signs alternate, because odd powers of a negative number are negative and even powers are positive.
• Symmetry: the numbers multiplying the terms (the coefficients) read the same forwards and backwards. These are studied in 3.1.2.

Worked examples

Example 1: Write down the powers that appear in the terms of (a + b)⁴.

Number of terms = n + 1 = 4 + 1 = 5

Terms follow the pattern: a⁴, a³b, a²b², ab³, b⁴

Answer: five terms, with a falling from power 4 to 0 and b rising from power 0 to 4, each pair adding up to 4.

Example 2: A 2 m flat bar is lengthened by 3% in a design revision at a fabrication workshop. Find the new length using (1 + x)^n with n = 1.

Formula: new length = L × (1 + x)^1

Substitution: new length = 2 × (1 + 0.03)

Working: 2 × 1.03 = 2.06

Answer: 2.06 m

Example 3: A square steel plate of side 1.5 m is made 0.03 m longer on each side. Find the factor by which its area increases, using (1 + x)².

x = 0.03 ÷ 1.5 = 0.02

Formula: (1 + x)² = 1 + 2x + x²

Substitution: 1 + 2 × 0.02 + (0.02)²

Working: 1 + 0.04 + 0.0004 = 1.0404

Answer: the area increases by a factor of 1.0404, which is an increase of 4.04%.

3.1.2 Coefficients

The coefficients of a binomial expansion are the numbers that multiply each term. The coefficient of the term containing b^k in the expansion of (a + b)^n is called the binomial coefficient, written C(n, k) and read "n choose k". In (a + b)³ = a³ + 3a²b + 3ab² + b³, the coefficients are 1, 3, 3 and 1, which are C(3, 0), C(3, 1), C(3, 2) and C(3, 3).

The binomial coefficient is found from factorials, where k! = k × (k − 1) × ... × 2 × 1 and 0! = 1:

C(n, k) = n! ÷ (k! × (n − k)!)

An equivalent form, which is quicker when k is small, is:

C(n, k) = n × (n − 1) × (n − 2) × ... × (n − k + 1) ÷ k!

The numerator has k factors, starting at n and falling by 1 each time.

Useful properties of binomial coefficients:

• Ends equal 1: C(n, 0) = 1 and C(n, n) = 1, so the first and last terms of every expansion have coefficient 1.
• Second coefficient: C(n, 1) = n and C(n, n − 1) = n, so the second and second-last coefficients equal the power.
• Symmetry: C(n, k) = C(n, n − k). For example C(7, 2) = C(7, 5) = 21, so only half of the coefficients need to be calculated.
• Sum of coefficients: the coefficients of (a + b)^n add up to 2^n. For n = 4: 1 + 4 + 6 + 4 + 1 = 16 = 2⁴.
• Addition rule: C(n, k) = C(n − 1, k − 1) + C(n − 1, k). This rule is the basis of Pascal's triangle in 3.1.3.

Worked examples

Example 1: Calculate C(5, 2).

Formula: C(n, k) = n × (n − 1) × ... × (n − k + 1) ÷ k!

Substitution: C(5, 2) = (5 × 4) ÷ (2 × 1)

Working: 20 ÷ 2 = 10

Answer: 10

Example 2: Calculate C(6, 3).

Substitution: C(6, 3) = (6 × 5 × 4) ÷ (3 × 2 × 1)

Working: 120 ÷ 6 = 20

Answer: 20

Example 3: Calculate C(8, 6) using symmetry.

C(8, 6) = C(8, 2)

Substitution: C(8, 2) = (8 × 7) ÷ (2 × 1)

Working: 56 ÷ 2 = 28

Answer: 28

Example 4: Calculate C(7, 4) using the factorial formula.

Formula: C(n, k) = n! ÷ (k! × (n − k)!)

Substitution: C(7, 4) = 7! ÷ (4! × 3!)

Working: 5040 ÷ (24 × 6) = 5040 ÷ 144 = 35

Answer: 35

3.1.3 Pascal's triangle

Pascal's triangle is a triangular arrangement of numbers in which each row lists the binomial coefficients for one power. It lets you write the coefficients of a small power such as (a + b)⁵ without calculating any factorials. Row 0 corresponds to the power 0, row 1 to the power 1, and so on, so row n contains the n + 1 coefficients C(n, 0), C(n, 1), ..., C(n, n).

Building the triangle:

1. Start the top: write 1 as row 0. This is the coefficient of (a + b)⁰ = 1.
2. Start each row with 1: every row begins and ends with 1, because C(n, 0) = C(n, n) = 1.
3. Add neighbours: each number inside a row is the sum of the two numbers directly above it, to the left and right. This is the addition rule C(n, k) = C(n − 1, k − 1) + C(n − 1, k).
4. Fill the row: work from the left until the row has one more number than the row above it.
5. Check symmetry: each row should read the same from left to right and right to left. If it does not, an addition error has been made.
6. Check the sum: the numbers in row n should add up to 2^n. Row 5 gives 1 + 5 + 10 + 10 + 5 + 1 = 32 = 2⁵.

The first rows of the triangle are:

• Row 0: 1
• Row 1: 1 1
• Row 2: 1 2 1
• Row 3: 1 3 3 1
• Row 4: 1 4 6 4 1
• Row 5: 1 5 10 10 5 1
• Row 6: 1 6 15 20 15 6 1
• Row 7: 1 7 21 35 35 21 7 1

Pascal's triangle is practical up to about row 10. For larger powers, or a single term from the middle of a large expansion, use the formula for C(n, k) from 3.1.2.

Worked examples

Example 1: Use Pascal's triangle to find row 8.

Row 7 is 1 7 21 35 35 21 7 1

Add neighbouring pairs: 1 + 7 = 8, 7 + 21 = 28, 21 + 35 = 56, 35 + 35 = 70, then 56, 28, 8 by symmetry

Answer: Row 8 is 1 8 28 56 70 56 28 8 1

Example 2: Use row 4 to expand (x + 2)⁴.

Coefficients from row 4: 1, 4, 6, 4, 1

Working: 1·x⁴ + 4·x³·2 + 6·x²·2² + 4·x·2³ + 1·2⁴

= x⁴ + 8x³ + 24x² + 32x + 16

Answer: x⁴ + 8x³ + 24x² + 32x + 16

Example 3: Check that the coefficients in row 6 agree with the formula for C(6, 2).

Pascal's triangle: the third number in row 6 is 15

Formula: C(6, 2) = (6 × 5) ÷ (2 × 1) = 30 ÷ 2 = 15

Answer: both methods give 15.

3.1.4 Expansion

To expand a power of a binomial, combine the binomial coefficients with the falling powers of the first term and the rising powers of the second term. For a positive whole number n, the binomial expansion is:

(a + b)^n = C(n, 0)a^n + C(n, 1)a^(n−1)b + C(n, 2)a^(n−2)b² + ... + C(n, n)b^n

When the first term is 1, the expansion in powers of x becomes:

(1 + x)^n = 1 + n x + n(n − 1)/2! × x² + n(n − 1)(n − 2)/3! × x³ + ... + x^n

For a positive whole number n, this series stops after n + 1 terms and is exact. When x is small, the first few terms already give a good approximation, because x², x³ and higher powers become very small. Welding engineers use this to estimate changes in dimensions, areas and volumes without a calculator.

Procedure for expanding (a + b)^n:

1. Identify a, b and n: write the first term, the second term (with its sign) and the power. In (2x − 1)³, a = 2x, b = −1 and n = 3.
2. Find the coefficients: read row n of Pascal's triangle, or calculate C(n, k) for k = 0 to n. For n = 3 the coefficients are 1, 3, 3, 1.
3. Write the powers: list a^n, a^(n−1)b, ..., b^n, keeping the brackets around a and b so that signs and numerical factors are raised to the power correctly.
4. Multiply each term: multiply each coefficient by its pair of powers.
5. Simplify: evaluate the numerical powers and multiply out each term.
6. Check the result: count the terms (n + 1), check the signs, and substitute a simple value such as x = 1 into both the original and the expansion to confirm they agree.

Worked examples

Example 1: Expand (1 + x)⁵.

Coefficients from row 5: 1, 5, 10, 10, 5, 1

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

Answer: 1 + 5x + 10x² + 10x³ + 5x⁴ + x⁵

Example 2: Expand (2x + 3)⁴.

Coefficients: 1, 4, 6, 4, 1

Working: (2x)⁴ + 4(2x)³(3) + 6(2x)²(3)² + 4(2x)(3)³ + 3⁴

= 16x⁴ + 4(8x³)(3) + 6(4x²)(9) + 4(2x)(27) + 81

= 16x⁴ + 96x³ + 216x² + 216x + 81

Answer: 16x⁴ + 96x³ + 216x² + 216x + 81

Example 3: Expand (x − 2)⁵.

Coefficients: 1, 5, 10, 10, 5, 1, with b = −2

Working: x⁵ + 5x⁴(−2) + 10x³(−2)² + 10x²(−2)³ + 5x(−2)⁴ + (−2)⁵

= x⁵ − 10x⁴ + 40x³ − 80x² + 80x − 32

Answer: x⁵ − 10x⁴ + 40x³ − 80x² + 80x − 32

Example 4: Expand (2x − 1)³.

Coefficients: 1, 3, 3, 1

Working: (2x)³ + 3(2x)²(−1) + 3(2x)(−1)² + (−1)³

= 8x³ − 12x² + 6x − 1

Answer: 8x³ − 12x² + 6x − 1

Example 5: Approximate (1 + 0.1)² using the first three terms of the series.

Given: n = 2, x = 0.1

Formula: (1 + x)^n = 1 + n x + n(n − 1)/2 × x²

Substitution: 1 + 2 × 0.1 + (2 × 1)/2 × (0.1)²

Working: 1 + 0.2 + 0.01 = 1.21

Answer: 1.21

Example 6: Use the first four terms to approximate (1.02)⁵, for example the scaling factor for a part whose dimensions are increased by 2% five times in succession.

Given: n = 5, x = 0.02

Formula: (1 + x)^5 = 1 + 5x + 10x² + 10x³ + ...

Substitution: 1 + 5(0.02) + 10(0.02)² + 10(0.02)³

Working: 1 + 0.1 + 10(0.0004) + 10(0.000008) = 1 + 0.1 + 0.004 + 0.00008

Answer: 1.10408

Example 7: Evaluate 99³ using the expansion of (100 − 1)³.

Formula: (a + b)³ = a³ + 3a²b + 3ab² + b³ with a = 100, b = −1

Working: 100³ + 3(100)²(−1) + 3(100)(−1)² + (−1)³

= 1 000 000 − 30 000 + 300 − 1

Answer: 970 299

Example 8: The radius of a steel pipe of radius 0.5 m increases by 0.02 m. Find the increase in cross-sectional area.

Given: r = 0.5 m, x = 0.02 ÷ 0.5 = 0.04

Area factor: (1 + 0.04)² = 1 + 0.08 + 0.0016 = 1.0816

Original area: A = π × r² = π × (0.5)² = 0.7854 m²

New area: 1.0816 × 0.7854 = 0.8495 m²

Increase: 0.8495 − 0.7854 = 0.0641 m²

Answer: the area increases by 0.0641 m²

Example 9: The radius of a cylindrical weld bead increases by 5% and its height by 2%. The volume is proportional to r² h. Find the percentage increase in volume.

Given: x_r = 0.05, x_h = 0.02

Formula: volume factor = (1 + x_r)² × (1 + x_h)

Substitution: (1 + 2 × 0.05 + 0.05²) × (1 + 0.02)

Working: (1 + 0.1 + 0.0025) × 1.02 = 1.1025 × 1.02 = 1.12455

Answer: the volume increases by about 12.5% (factor 1.12455).

Example 10: A cubic steel test block has side length increased by 3%. Its volume is proportional to (side)³. Find the percentage increase in volume using the expansion up to x².

Given: x = 0.03, n = 3

Formula: (1 + x)³ ≈ 1 + 3x + 3x²

Substitution: 1 + 3(0.03) + 3(0.03)²

Working: 1 + 0.09 + 0.0027 = 1.0927

Answer: the volume increases by about 9.27%. The omitted x³ term is only 0.000027, so the approximation is very close to the exact factor 1.092727.

Chapter Summary

This chapter introduced the concept of the binomial series, starting with the understanding of powers and how they relate to binomial expressions. It covered the importance of coefficients in the expansion process and demonstrated how Pascal's triangle serves as a practical tool for identifying these coefficients efficiently. The chapter then detailed the method of expanding binomial expressions using these coefficients. Moving into the binomial theorem, it explained the expansion for positive integer powers of \(n\), providing a foundation for more complex cases. The theorem was extended to include negative powers of \(n\), allowing for the expansion of expressions with inverse powers. Fractional powers of \(n\) were also explored, highlighting how roots can be handled within the binomial framework. Finally, the chapter addressed the estimation of errors associated with small changes in the binomial expansion, equipping students with techniques to assess the accuracy of approximations in practical engineering calculations.