Binomial expansion is an essential mathematical tool for an Auto Body Technician when calculating forces, stresses, or measurements that involve powers of binomials. Understanding how to expand powers of expressions such as \((a + b)^n\) helps in simplifying complex calculations during vehicle body repairs or modifications. This chapter focuses on the binomial series expansion, including how to determine powers, coefficients, and how to use Pascal’s triangle for quick expansions relevant to practical auto body work.

3.1 Binomial series expansion

Binomial series expansion allows the expression of powers of binomials in a simplified polynomial form. This is crucial in estimating values and solving engineering problems related to auto body structures, where approximations and expansions are often necessary.

3.1.1 Powers

The power \(n\) in a binomial expression \((a + b)^n\) determines how many terms will appear in the expanded form and the complexity of the calculation. The expansion results in \(n+1\) terms, each involving powers of \(a\) and \(b\) that sum to \(n\).

The binomial theorem states:$$ (a + b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^k $$ where \(\binom{n}{k}\) are the binomial coefficients.

Worked Examples

Example 1: Expand \((x + 2)^2\).

Given: \(n=2, a = x, b = 2\)

$$ (x + 2)^2 = \sum_{k=0}^2 \binom{2}{k} x^{2-k} 2^k $$

Calculate terms:

$$ \binom{2}{0} x^{2} 2^{0} = 1 \times x^{2} \times 1 = x^{2} $$

$$ \binom{2}{1} x^{1} 2^{1} = 2 \times x \times 2 = 4x $$

$$ \binom{2}{2} x^{0} 2^{2} = 1 \times 1 \times 4 = 4 $$

Answer:

$$ (x + 2)^2 = x^{2} + 4x + 4 $$

Example 2: Expand \((3 + y)^3\).

Given: \(n=3, a=3, b=y\)

$$ (3 + y)^3 = \sum_{k=0}^3 \binom{3}{k} 3^{3-k} y^k $$

Calculate terms:

$$ \binom{3}{0} 3^{3} y^{0} = 1 \times 27 \times 1 = 27 $$

$$ \binom{3}{1} 3^{2} y^{1} = 3 \times 9 \times y = 27y $$

$$ \binom{3}{2} 3^{1} y^{2} = 3 \times 3 \times y^{2} = 9y^{2} $$

$$ \binom{3}{3} 3^{0} y^{3} = 1 \times 1 \times y^{3} = y^{3} $$

Answer:

$$ (3 + y)^3 = 27 + 27y + 9y^{2} + y^{3} $$

Example 3: Expand \((2x + 5)^4\).

Given: \(n=4, a=2x, b=5\)

$$ (2x + 5)^4 = \sum_{k=0}^4 \binom{4}{k} (2x)^{4-k} 5^k $$

Calculate terms:

$$ \binom{4}{0} (2x)^4 5^0 = 1 \times 16x^{4} \times 1 = 16x^{4} $$

$$ \binom{4}{1} (2x)^3 5^1 = 4 \times 8x^{3} \times 5 = 160x^{3} $$

$$ \binom{4}{2} (2x)^2 5^2 = 6 \times 4x^{2} \times 25 = 600x^{2} $$

$$ \binom{4}{3} (2x)^1 5^3 = 4 \times 2x \times 125 = 1000x $$

$$ \binom{4}{4} (2x)^0 5^4 = 1 \times 1 \times 625 = 625 $$

Answer:

$$ (2x + 5)^4 = 16x^{4} + 160x^{3} + 600x^{2} + 1000x + 625 $$

3.1.2 Coefficients

Binomial coefficients \(\binom{n}{k}\) determine the multiplier for each term in the binomial expansion. They are calculated using factorials:

$$ \binom{n}{k} = \frac{n!}{k!(n-k)!} $$

These coefficients indicate how many ways to choose \(k\) items from \(n\) and are key in determining term weights in the expansion.

Worked Examples

Example 1: Calculate \(\binom{5}{2}\).

Given: \(n=5, k=2\)

$$ \binom{5}{2} = \frac{5!}{2! \times 3!} = \frac{120}{2 \times 6} = \frac{120}{12} = 10 $$

Answer:

$$ \binom{5}{2} = 10 $$

Example 2: Calculate \(\binom{6}{4}\).

Given: \(n=6, k=4\)

$$ \binom{6}{4} = \frac{6!}{4! \times 2!} = \frac{720}{24 \times 2} = \frac{720}{48} = 15 $$

Answer:

$$ \binom{6}{4} = 15 $$

Example 3: Calculate \(\binom{7}{0}\).

Given: \(n=7, k=0\)

$$ \binom{7}{0} = \frac{7!}{0! \times 7!} = \frac{5040}{1 \times 5040} = 1 $$

Answer:

$$ \binom{7}{0} = 1 $$

Example 4: Calculate \(\binom{8}{3}\).

Given: \(n=8, k=3\)

$$ \binom{8}{3} = \frac{8!}{3! \times 5!} = \frac{40320}{6 \times 120} = \frac{40320}{720} = 56 $$

Answer:

$$ \binom{8}{3} = 56 $$

Example 5: Calculate \(\binom{9}{5}\).

Given: \(n=9, k=5\)

$$ \binom{9}{5} = \frac{9!}{5! \times 4!} = \frac{362880}{120 \times 24} = \frac{362880}{2880} = 126 $$

Answer:

$$ \binom{9}{5} = 126 $$

3.1.3 Pascal's triangle

Pascal’s triangle is a triangular array of binomial coefficients that provides a quick reference for coefficients in binomial expansions without factorial calculations. Each number is the sum of the two numbers directly above it.

The triangle starts as:

\[ \begin{array}{cccccc} & & 1 & & \\ & 1 & & 1 & \\ 1 & & 2 & & 1 \\ & 1 & & 3 & & 3 & & 1 \\ 1 & & 4 & & 6 & & 4 & & 1 \\ \end{array} \]

Worked Examples

Example 1: Find coefficients for \((a + b)^4\) using Pascal’s triangle.

From the fifth row (starting count at zero), coefficients are: 1, 4, 6, 4, 1.

Answer:

Coefficients are 1, 4, 6, 4, 1.

Example 2: Use Pascal’s triangle to expand \((x + 3)^3\).

Coefficients from the fourth row: 1, 3, 3, 1.

Expansion:

$$ x^{3} + 3 \times x^{2} \times 3 + 3 \times x \times 3^{2} + 3^{3} $$

Calculate terms:

$$ x^{3} + 9x^{2} + 27x + 27 $$

Answer:

$$ (x + 3)^3 = x^{3} + 9x^{2} + 27x + 27 $$

Example 3: Find the 7th row of Pascal’s triangle and use it for \((a + b)^6\).

Seventh row: 1, 6, 15, 20, 15, 6, 1.

Expansion:

$$ a^{6} + 6a^{5}b + 15a^{4}b^{2} + 20a^{3}b^{3} + 15a^{2}b^{4} + 6ab^{5} + b^{6} $$

Answer:

Coefficients for \((a + b)^6\) are 1, 6, 15, 20, 15, 6, 1.

Example 4: Using Pascal’s triangle, write the coefficients for \((p + q)^5\).

Row six coefficients: 1, 5, 10, 10, 5, 1.

Answer:

Coefficients are 1, 5, 10, 10, 5, 1.

3.1.4 Expansion

Expanding a binomial expression involves applying the binomial theorem or using Pascal’s triangle to write out all terms explicitly. Each term is formed by multiplying the binomial coefficient, the power of the first term, and the power of the second term.

The general term \(T_{k+1}\) in the expansion of \((a + b)^n\) is:

$$ T_{k+1} = \binom{n}{k} a^{n-k} b^k $$

where \(k = 0, 1, 2, \ldots, n\).

Worked Examples

Example 1: Expand \((x + 1)^3\) fully.

Given: \(n=3, a=x, b=1\)

Terms:

$$ T_1 = \binom{3}{0} x^{3} 1^{0} = 1 \times x^{3} \times 1 = x^{3} $$

$$ T_2 = \binom{3}{1} x^{2} 1^{1} = 3 \times x^{2} \times 1 = 3x^{2} $$

$$ T_3 = \binom{3}{2} x^{1} 1^{2} = 3 \times x \times 1 = 3x $$

$$ T_4 = \binom{3}{3} x^{0} 1^{3} = 1 \times 1 \times 1 = 1 $$

Answer:

$$ (x + 1)^3 = x^{3} + 3x^{2} + 3x + 1 $$

Example 2: Expand \((2 + y)^4\).

Given: \(n=4, a=2, b=y\)

Terms:

$$ T_1 = \binom{4}{0} 2^{4} y^{0} = 1 \times 16 \times 1 = 16 $$

$$ T_2 = \binom{4}{1} 2^{3} y^{1} = 4 \times 8 \times y = 32y $$

$$ T_3 = \binom{4}{2} 2^{2} y^{2} = 6 \times 4 \times y^{2} = 24y^{2} $$

$$ T_4 = \binom{4}{3} 2^{1} y^{3} = 4 \times 2 \times y^{3} = 8y^{3} $$

$$ T_5 = \binom{4}{4} 2^{0} y^{4} = 1 \times 1 \times y^{4} = y^{4} $$

Answer:

$$ (2 + y)^4 = 16 + 32y + 24y^{2} + 8y^{3} + y^{4} $$

Example 3: Expand \((x + 3)^5\).

Given: \(n=5, a=x, b=3\)

Terms:

$$ T_1 = \binom{5}{0} x^{5} 3^{0} = 1 \times x^{5} \times 1 = x^{5} $$

$$ T_2 = \binom{5}{1} x^{4} 3^{1} = 5 \times x^{4} \times 3 = 15x^{4} $$

$$ T_3 = \binom{5}{2} x^{3} 3^{2} = 10 \times x^{3} \times 9 = 90x^{3} $$

$$ T_4 = \binom{5}{3} x^{2} 3^{3} = 10 \times x^{2} \times 27 = 270x^{2} $$

$$ T_5 = \binom{5}{4} x^{1} 3^{4} = 5 \times x \times 81 = 405x $$

$$ T_6 = \binom{5}{5} x^{0} 3^{5} = 1 \times 1 \times 243 = 243 $$

Answer:

$$ (x + 3)^5 = x^{5} + 15x^{4} + 90x^{3} + 270x^{2} + 405x + 243 $$

Example 4: Expand \((4x + 2)^3\).

Given: \(n=3, a=4x, b=2\)

Terms:

$$ T_1 = \binom{3}{0} (4x)^3 2^{0} = 1 \times 64x^{3} \times 1 = 64x^{3} $$

$$ T_2 = \binom{3}{1} (4x)^2 2^{1} = 3 \times 16x^{2} \times 2 = 96x^{2} $$

$$ T_3 = \binom{3}{2} (4x)^1 2^{2} = 3 \times 4x \times 4 = 48x $$

$$ T_4 = \binom{3}{3} (4x)^0 2^{3} = 1 \times 1 \times 8 = 8 $$

Answer:

$$ (4x + 2)^3 = 64x^{3} + 96x^{2} + 48x + 8 $$

Example 5: Expand \((x + y)^6\).

Given: \(n=6, a=x, b=y\)

Coefficients from Pascal’s triangle: 1, 6, 15, 20, 15, 6, 1

Terms:

$$ T_1 = 1 \times x^{6} \times y^{0} = x^{6} $$

$$ T_2 = 6 \times x^{5} \times y^{1} = 6x^{5}y $$

$$ T_3 = 15 \times x^{4} \times y^{2} = 15x^{4}y^{2} $$

$$ T_4 = 20 \times x^{3} \times y^{3} = 20x^{3}y^{3} $$

$$ T_5 = 15 \times x^{2} \times y^{4} = 15x^{2}y^{4} $$

$$ T_6 = 6 \times x^{1} \times y^{5} = 6xy^{5} $$

$$ T_7 = 1 \times x^{0} \times y^{6} = y^{6} $$

Answer:

$$ (x + y)^6 = x^{6} + 6x^{5}y + 15x^{4}y^{2} + 20x^{3}y^{3} + 15x^{2}y^{4} + 6xy^{5} + y^{6} $$

Chapter Summary

This chapter focused on the binomial expansion, beginning with the binomial series expansion where powers, coefficients, and Pascal's triangle were introduced as essential tools for understanding the structure of the expansion. The method of expanding expressions using binomial coefficients was demonstrated, emphasizing the role of Pascal's triangle in determining these coefficients efficiently. The binomial theorem was then explored in greater depth, first addressing expansions with positive integer powers of \(n\). The study progressed to handling negative powers and fractional powers of \(n\), which require more advanced techniques due to the infinite nature of their series. The chapter concluded by examining how to estimate errors resulting from small changes in the variables within the binomial expansion, an important aspect for practical engineering calculations. Throughout, the focus remained on applying these concepts to relevant civil engineering problems, ensuring a strong foundation in both theory and application.