By the end of this chapter, you will be able to:
Mastering these skills will help you solve practical engineering problems with precision and confidence in your trade.
Binomial expansion is a fundamental mathematical tool for chemical engineers in Kenya, enabling them to approximate complex expressions encountered in process calculations, reaction kinetics, and fluid mechanics. Mastery of binomial expansion facilitates efficient estimation of small changes and error analysis in systems such as reactor design or separation processes. This chapter covers Pascal’s triangle, the binomial theorem, root approximations using binomial expansions, error estimation for small perturbations, and the role of factorials and combinations in the theorem.
Pascal’s triangle is a triangular array of coefficients that arise in the expansion of binomial expressions. Each number in the triangle is the sum of the two numbers directly above it, starting with 1 at the apex. This triangle provides binomial coefficients for expansions of the form \((a + b)^n\), which are essential in chemical engineering calculations such as polynomial approximations in thermodynamics or reaction rate expressions.
The coefficients for \((a + b)^n\) are given by the \(n^{th}\) row of Pascal’s triangle, starting from row zero.
$$ \begin{array}{cccccc} n=0: & 1 \\ n=1: & 1 & 1 \\ n=2: & 1 & 2 & 1 \\ n=3: & 1 & 3 & 3 & 1 \\ n=4: & 1 & 4 & 6 & 4 & 1 \\ n=5: & 1 & 5 & 10 & 10 & 5 & 1 \\ n=6: & 1 & 6 & 15 & 20 & 15 & 6 & 1 \\ \end{array} $$
Example 1: Find the coefficients of \((a + b)^4\) using Pascal’s triangle.
Given: \(n = 4\)
The 4th row of Pascal’s triangle is: 1, 4, 6, 4, 1
Answer: The coefficients are 1, 4, 6, 4, 1.
Example 2: Determine the coefficient of the term with \(a^3 b^2\) in \((a + b)^5\).
Given: \(n = 5\), term corresponds to powers \(a^3 b^2\)
From Pascal’s triangle row 5: 1, 5, 10, 10, 5, 1
The term with \(a^{n-k} b^k\) has coefficient \(\binom{n}{k}\), here \(k=2\)
Coefficient = 10
Answer: The coefficient is 10.
Example 3: Using Pascal’s triangle, write the full expansion of \((x + y)^3\).
Coefficients from row 3: 1, 3, 3, 1
Expansion:
$$ (x + y)^3 = 1 \cdot x^3 + 3 \cdot x^2 y + 3 \cdot x y^2 + 1 \cdot y^3 $$
Answer: \(x^3 + 3 x^2 y + 3 x y^2 + y^3\)
Example 4: Find the coefficient of the middle term in \((2a + b)^4\) using Pascal’s triangle.
Row 4 coefficients: 1, 4, 6, 4, 1
The middle term corresponds to the term with coefficient 6.
Answer: The middle term coefficient is 6.
Example 5: Verify that the sum of the coefficients in the expansion of \((a + b)^6\) equals \(2^6\).
Coefficients: 1, 6, 15, 20, 15, 6, 1
Sum: \(1 + 6 + 15 + 20 + 15 + 6 + 1 = 64\)
Calculate \(2^6 = 64\)
Answer: Sum of coefficients is 64, which equals \(2^6\).
The binomial theorem provides a general formula to expand expressions of the form \((a + b)^n\) for any non-negative integer \(n\). It expresses the expansion as a sum of terms involving binomial coefficients, powers of \(a\), and powers of \(b\). This theorem is crucial in chemical engineering for approximating nonlinear terms in process equations and reaction kinetics.
The binomial theorem states:
$$ (a + b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} b^k $$
where \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\) are binomial coefficients.
Example 1: Expand \((x + 2)^3\) using the binomial theorem.
Given: \(a = x\), \(b = 2\), \(n = 3\)
Formula:
$$ (x + 2)^3 = \sum_{k=0}^3 \binom{3}{k} x^{3-k} 2^k $$
Calculate terms:
For \(k=0\): \(\binom{3}{0} x^3 2^0 = 1 \times x^3 \times 1 = x^3\)
For \(k=1\): \(\binom{3}{1} x^2 2^1 = 3 \times x^2 \times 2 = 6x^2\)
For \(k=2\): \(\binom{3}{2} x^1 2^2 = 3 \times x \times 4 = 12x\)
For \(k=3\): \(\binom{3}{3} x^0 2^3 = 1 \times 1 \times 8 = 8\)
Sum terms:
$$ x^3 + 6x^2 + 12x + 8 $$
Answer: \(x^3 + 6x^2 + 12x + 8\)
Example 2: Expand \((3 + y)^4\).
Given: \(a = 3\), \(b = y\), \(n = 4\)
Formula:
$$ (3 + y)^4 = \sum_{k=0}^4 \binom{4}{k} 3^{4-k} y^k $$
Calculate terms:
\(k=0\): \(1 \times 3^4 \times y^0 = 81\)
\(k=1\): \(4 \times 3^3 \times y = 4 \times 27 \times y = 108 y\)
\(k=2\): \(6 \times 3^2 \times y^2 = 6 \times 9 \times y^2 = 54 y^2\)
\(k=3\): \(4 \times 3^1 \times y^3 = 4 \times 3 \times y^3 = 12 y^3\)
\(k=4\): \(1 \times 3^0 \times y^4 = y^4\)
Sum terms:
$$ 81 + 108 y + 54 y^2 + 12 y^3 + y^4 $$
Answer: \(81 + 108 y + 54 y^2 + 12 y^3 + y^4\)
Example 3: Expand \((2m - 1)^3\).
Rewrite as \((a + b)^3\) with \(a=2m\), \(b = -1\).
Formula:
$$ (2m - 1)^3 = \sum_{k=0}^3 \binom{3}{k} (2m)^{3-k} (-1)^k $$
Calculate terms:
\(k=0\): \(1 \times (2m)^3 \times 1 = 8 m^3\)
\(k=1\): \(3 \times (2m)^2 \times (-1) = 3 \times 4 m^2 \times (-1) = -12 m^2\)
\(k=2\): \(3 \times (2m)^1 \times 1 = 3 \times 2 m \times 1 = 6 m\)
\(k=3\): \(1 \times (2m)^0 \times (-1)^3 = 1 \times 1 \times (-1) = -1\)
Sum terms:
$$ 8 m^3 - 12 m^2 + 6 m - 1 $$
Answer: \(8 m^3 - 12 m^2 + 6 m - 1\)
Example 4: Expand \((1 + \frac{x}{2})^5\).
Given: \(a = 1\), \(b = \frac{x}{2}\), \(n = 5\)
Formula:
$$ (1 + \frac{x}{2})^5 = \sum_{k=0}^5 \binom{5}{k} 1^{5-k} (\frac{x}{2})^k $$
Calculate terms:
\(k=0\): \(1 \times 1 \times 1 = 1\)
\(k=1\): \(5 \times 1 \times \frac{x}{2} = \frac{5x}{2}\)
\(k=2\): \(10 \times 1 \times \frac{x^2}{4} = \frac{10 x^2}{4} = \frac{5 x^2}{2}\)
\(k=3\): \(10 \times 1 \times \frac{x^3}{8} = \frac{10 x^3}{8} = \frac{5 x^3}{4}\)
\(k=4\): \(5 \times 1 \times \frac{x^4}{16} = \frac{5 x^4}{16}\)
\(k=5\): \(1 \times 1 \times \frac{x^5}{32} = \frac{x^5}{32}\)
Sum terms:
$$ 1 + \frac{5x}{2} + \frac{5 x^2}{2} + \frac{5 x^3}{4} + \frac{5 x^4}{16} + \frac{x^5}{32} $$
Answer: \(1 + \frac{5x}{2} + \frac{5 x^2}{2} + \frac{5 x^3}{4} + \frac{5 x^4}{16} + \frac{x^5}{32}\)
Example 5: Expand \((1 - \frac{3y}{4})^6\).
Given: \(a = 1\), \(b = -\frac{3y}{4}\), \(n = 6\)
Formula:
$$ (1 - \frac{3y}{4})^6 = \sum_{k=0}^6 \binom{6}{k} 1^{6-k} (-\frac{3y}{4})^k $$
Calculate terms:
\(k=0\): \(1\)
\(k=1\): \(6 \times (-\frac{3y}{4}) = -\frac{18 y}{4} = -\frac{9 y}{2}\)
\(k=2\): \(15 \times (\frac{9 y^2}{16}) = \frac{135 y^2}{16}\)
\(k=3\): \(20 \times (-\frac{27 y^3}{64}) = -\frac{540 y^3}{64} = -\frac{135 y^3}{16}\)
\(k=4\): \(15 \times \frac{81 y^4}{256} = \frac{1215 y^4}{256}\)
\(k=5\): \(6 \times (-\frac{243 y^5}{1024}) = -\frac{1458 y^5}{1024}\)
\(k=6\): \(1 \times \frac{729 y^6}{4096} = \frac{729 y^6}{4096}\)
Sum terms:
$$ 1 - \frac{9 y}{2} + \frac{135 y^2}{16} - \frac{135 y^3}{16} + \frac{1215 y^4}{256} - \frac{1458 y^5}{1024} + \frac{729 y^6}{4096} $$
Answer: \(1 - \frac{9 y}{2} + \frac{135 y^2}{16} - \frac{135 y^3}{16} + \frac{1215 y^4}{256} - \frac{1458 y^5}{1024} + \frac{729 y^6}{4096}\)
Example 6: A process engineer at Bamburi Cement needs to expand \((4 + x)^3\) for process calculations.
Given: \(a = 4\), \(b = x\), \(n = 3\)
Formula:$$ (4 + x)^3 = \sum_{k=0}^3 \binom{3}{k} 4^{3-k} x^k $$ Calculate terms:
For \(k=0\): \(\binom{3}{0} 4^3 x^0 = 1 \times 64 \times 1 = 64\)
For \(k=1\): \(\binom{3}{1} 4^2 x = 3 \times 16 \times x = 48x\)
For \(k=2\): \(\binom{3}{2} 4^1 x^2 = 3 \times 4 \times x^2 = 12x^2\)
For \(k=3\): \(\binom{3}{3} 4^0 x^3 = 1 \times 1 \times x^3 = x^3\)
Sum terms:$$ 64 + 48x + 12x^2 + x^3 $$
Answer: \(64 + 48x + 12x^2 + x^3\)
Example 7: A technician at Kenya Bureau of Standards needs to expand \((1 - 2y)^4\) for quality control calculations.
Given: \(a = 1\), \(b = -2y\), \(n = 4\)
Formula:$$ (1 - 2y)^4 = \sum_{k=0}^4 \binom{4}{k} 1^{4-k} (-2y)^k $$ Calculate terms:
\(k=0\): \(1\)
\(k=1\): \(4 \times (-2y) = -8y\)
\(k=2\): \(6 \times (4y^2) = 24y^2\)
\(k=3\): \(4 \times (-8y^3) = -32y^3\)
\(k=4\): \(1 \times 16y^4 = 16y^4\)
Sum terms:$$ 1 - 8y + 24y^2 - 32y^3 + 16y^4 $$
Answer: \(1 - 8y + 24y^2 - 32y^3 + 16y^4\)
The binomial theorem can approximate roots of numbers by expressing the root as a binomial expression with a small perturbation. This method is particularly useful in chemical engineering when calculating non-integer roots, such as square roots or cube roots of process variables, without a calculator.
For \(|x| < 1\), the expansion:
$$ (1 + x)^n = 1 + n x + \frac{n(n-1)}{2!} x^2 + \frac{n(n-1)(n-2)}{3!} x^3 + \cdots $$
can be used to approximate roots by setting \(n = \frac{1}{m}\) for the \(m^{th}\) root.
Example 1: Approximate \(\sqrt{1.04}\) using binomial expansion.
Rewrite as:
$$ \sqrt{1.04} = (1 + 0.04)^{\frac{1}{2}} $$
Using expansion up to second order:
$$ (1 + x)^n \approx 1 + n x + \frac{n(n-1)}{2} x^2 $$
Given: \(x = 0.04\), \(n = \frac{1}{2}\)
Calculate:
\(1 + \frac{1}{2} \times 0.04 + \frac{\frac{1}{2} \times (\frac{1}{2} - 1)}{2} \times (0.04)^2\)
First term: \(1\)
Second term: \(\frac{1}{2} \times 0.04 = 0.02\)
Third term:
$$ \frac{\frac{1}{2} \times (-\frac{1}{2})}{2} \times 0.0016 = \frac{-\frac{1}{4}}{2} \times 0.0016 = -\frac{1}{8} \times 0.0016 = -0.0002 $$
Sum:
$$ 1 + 0.02 - 0.0002 = 1.0198 $$
Answer: \(\sqrt{1.04} \approx 1.0198\)
Example 2: Approximate \(\sqrt[3]{0.992}\).
Rewrite as:
$$ (1 - 0.008)^{\frac{1}{3}} $$
Using first two terms:
$$ 1 + \frac{1}{3} (-0.008) + \frac{\frac{1}{3} (\frac{1}{3} - 1)}{2} (-0.008)^2 $$
Calculate:
Second term:
$$ \frac{1}{3} \times (-0.008) = -0.0026667 $$
Third term:
$$ \frac{\frac{1}{3} \times (-\frac{2}{3})}{2} \times 0.000064 = \frac{-\frac{2}{9}}{2} \times 0.000064 = -\frac{1}{9} \times 0.000064 = -0.00000711 $$
Sum:
$$ 1 - 0.0026667 - 0.00000711 = 0.997326 $$
Answer: \(\sqrt[3]{0.992} \approx 0.9973\)
Example 3: Approximate \(\sqrt{9.2}\).
Rewrite as:
$$ \sqrt{9.2} = \sqrt{9 \times (1 + 0.0222)} = 3 \times (1 + 0.0222)^{\frac{1}{2}} $$
Use binomial expansion for \((1 + 0.0222)^{\frac{1}{2}}\):
$$ 1 + \frac{1}{2} \times 0.0222 + \frac{\frac{1}{2} \times (-\frac{1}{2})}{2} \times (0.0222)^2 $$
Calculate:
Second term: \(0.0111\)
Third term:
$$ \frac{-\frac{1}{4}}{2} \times 0.0004938 = -\frac{1}{8} \times 0.0004938 = -0.0000617 $$
Sum:
$$ 1 + 0.0111 - 0.0000617 = 1.01104 $$
Multiply by 3:
$$ 3 \times 1.01104 = 3.0331 $$
Answer: \(\sqrt{9.2} \approx 3.0331\)
Example 4: Approximate \(\sqrt[4]{16.5}\).
Rewrite as:
$$ \sqrt[4]{16.5} = \sqrt[4]{16 \times (1 + 0.03125)} = 2 \times (1 + 0.03125)^{\frac{1}{4}} $$
Use binomial expansion up to second term:
$$ 1 + \frac{1}{4} \times 0.03125 + \frac{\frac{1}{4} \times (\frac{1}{4} - 1)}{2} \times (0.03125)^2 $$
Calculate:
Second term: \(0.0078125\)
Third term:
$$ \frac{\frac{1}{4} \times (-\frac{3}{4})}{2} \times 0.0009766 = \frac{-\frac{3}{16}}{2} \times 0.0009766 = -\frac{3}{32} \times 0.0009766 = -0.0000915 $$
Sum:
$$ 1 + 0.0078125 - 0.0000915 = 1.007721 $$
Multiply by 2:
$$ 2 \times 1.007721 = 2.0154 $$
Answer: \(\sqrt[4]{16.5} \approx 2.0154\)
Example 5: Approximate \(\sqrt[3]{27.5}\).
Rewrite as:
$$ \sqrt[3]{27.5} = \sqrt[3]{27 \times (1 + 0.01852)} = 3 \times (1 + 0.01852)^{\frac{1}{3}} $$
Use binomial expansion:
$$ 1 + \frac{1}{3} \times 0.01852 + \frac{\frac{1}{3} \times (\frac{1}{3} - 1)}{2} \times (0.01852)^2 $$
Calculate:
Second term: \(0.006174\)
Third term:
$$ \frac{\frac{1}{3} \times (-\frac{2}{3})}{2} \times 0.0003429 = \frac{-\frac{2}{9}}{2} \times 0.0003429 = -\frac{1}{9} \times 0.0003429 = -0.0000381 $$
Sum:
$$ 1 + 0.006174 - 0.0000381 = 1.006136 $$
Multiply by 3:
$$ 3 \times 1.006136 = 3.0184 $$
Answer: \(\sqrt[3]{27.5} \approx 3.0184\)
Example 6: Approximate \(\sqrt{1.06}\) for a process at Kenya Medical Research Institute using binomial expansion.
Rewrite as:$$ \sqrt{1.06} = (1 + 0.06)^{\frac{1}{2}} $$ Using expansion up to second order:$$ (1 + x)^n \approx 1 + n x + \frac{n(n-1)}{2} x^2 $$ Given: \(x = 0.06\), \(n = \frac{1}{2}\)
First term: \(1\)
Second term: \(\frac{1}{2} \times 0.06 = 0.03\)
Third term:$$
\frac{\frac{1}{2} \times (-\frac{1}{2})}{2} \times (0.06)^2 = \frac{-\frac{1}{4}}{2} \times 0.0036 = -\frac{1}{8} \times 0.0036 = -0.00045
$$
Sum:$$
1 + 0.03 - 0.00045 = 1.02955
$$
Answer: \(\sqrt{1.06} \approx 1.0296\)
In chemical engineering, small changes in process variables can cause errors in calculated results. The binomial theorem allows estimation of such errors by approximating how a function changes with a small perturbation. This is useful in error analysis for measurements, process control, and sensitivity studies.
For a function expressed as \((1 + x)^n\) with \(|x| \ll 1\), the approximate change is:
$$ (1 + x)^n \approx 1 + n x + \frac{n(n-1)}{2} x^2 $$
The error due to the small change \(x\) can be estimated by the difference from 1.
Example 1: Estimate the fractional error in volume when the radius of a spherical tank increases by 2%.
Volume \(V \propto r^3\), so \(n=3\), \(x=0.02\)
Approximate fractional change:
$$ (1 + 0.02)^3 \approx 1 + 3 \times 0.02 + \frac{3 \times 2}{2} \times (0.02)^2 $$
Calculate:
First term: 1
Second term: \(3 \times 0.02 = 0.06\)
Third term:
$$ \frac{3 \times 2}{2} \times 0.0004 = 3 \times 0.0004 = 0.0012 $$
Sum:
$$ 1 + 0.06 + 0.0012 = 1.0612 $$
Fractional increase = 0.0612 or 6.12%
Answer: Volume increases approximately by 6.12%
Example 2: Estimate the error in the flow rate \(Q = A v\) when the diameter of a pipe increases by 1% and velocity remains constant.
Area \(A \propto d^2\), so \(n=2\), \(x=0.01\)
Approximate fractional change:
$$ (1 + 0.01)^2 \approx 1 + 2 \times 0.01 + \frac{2 \times 1}{2} \times (0.01)^2 $$
Calculate:
Second term: \(0.02\)
Third term:
$$ 1 \times 0.0001 = 0.0001 $$
Sum:
$$ 1 + 0.02 + 0.0001 = 1.0201 $$
Fractional increase = 0.0201 or 2.01%
Answer: Area increases approximately by 2.01%
Example 3: Estimate the percentage change in pressure drop \(\Delta P \propto L\) when pipe length increases by 5%.
Here, \(n=1\), \(x=0.05\)
Approximate fractional change:
$$ (1 + 0.05)^1 = 1 + 1 \times 0.05 = 1.05 $$
No second order term since \(n=1\).
Fractional increase = 0.05 or 5%
Answer: Pressure drop increases by 5%
Example 4: Estimate the error in reaction rate \(r \propto C^2\) if concentration measurement has a 3% error.
Here, \(n=2\), \(x=0.03\)
Approximate fractional change:
$$ (1 + 0.03)^2 \approx 1 + 2 \times 0.03 + \frac{2 \times 1}{2} \times (0.03)^2 $$
Calculate:
Second term: 0.06
Third term:
$$ 1 \times 0.0009 = 0.0009 $$
Sum:
$$ 1 + 0.06 + 0.0009 = 1.0609 $$
Fractional increase = 0.0609 or 6.09%
Answer: Reaction rate increases approximately by 6.09%
Example 5: Estimate the percentage error in heat transfer coefficient \(h \propto (T_s - T_\infty)^{0.8}\) if the temperature difference changes by 4%.
Here, \(n=0.8\), \(x=0.04\)
Approximate fractional change:
$$ (1 + 0.04)^{0.8} \approx 1 + 0.8 \times 0.04 + \frac{0.8 \times (-0.2)}{2} \times (0.04)^2 $$
Calculate:
Second term: \(0.032\)
Third term:
$$ \frac{0.8 \times (-0.2)}{2} \times 0.0016 = \frac{-0.16}{2} \times 0.0016 = -0.08 \times 0.0016 = -0.000128 $$
Sum:
$$ 1 + 0.032 - 0.000128 = 1.03187 $$
Fractional increase = 0.03187 or 3.19%
Answer: Heat transfer coefficient increases approximately by 3.19%
Example 6: Estimate the percentage error in the cross-sectional area of a water supply pipe at Nairobi Water and Sewerage Company if its diameter increases by 2.5%.
Area \(A \propto d^2\), so \(n=2\), \(x=0.025\)
Approximate fractional change:$$
(1 + 0.025)^2 \approx 1 + 2 \times 0.025 + \frac{2 \times 1}{2} \times (0.025)^2
$$
Calculate:
Second term: \(0.05\)
Third term:$$
1 \times 0.000625 = 0.000625
$$
Sum:$$
1 + 0.05 + 0.000625 = 1.050625
$$
Fractional increase = 0.050625 or 5.06%
Answer: Area increases approximately by 5.06%
Factorials and combinations are integral to the binomial theorem, providing the mathematical basis for binomial coefficients. Factorials (\(n!\)) represent the product of all positive integers up to \(n\), while combinations \(\binom{n}{k}\) count the number of ways to choose \(k\) items from \(n\) without regard to order. These concepts are vital in chemical engineering for statistical mechanics, process optimization, and combinatorial calculations.
The binomial coefficient is defined as:
$$ \binom{n}{k} = \frac{n!}{k! (n-k)!} $$
where
$$ n! = n \times (n-1) \times (n-2) \times \cdots \times 2 \times 1, \quad 0! = 1 $$
Example 1: Calculate \(5!\).
Formula:
$$ 5! = 5 \times 4 \times 3 \times 2 \times 1 $$
Calculate:
$$ 5 \times 4 = 20 $$
$$ 20 \times 3 = 60 $$
$$ 60 \times 2 = 120 $$
$$ 120 \times 1 = 120 $$
Answer: \(5! = 120\)
Example 2: Calculate \(\binom{6}{2}\).
Formula:
$$ \binom{6}{2} = \frac{6!}{2! \times 4!} $$
Calculate factorials:
\(6! = 720\), \(2! = 2\), \(4! = 24\)
Calculate:
$$ \frac{720}{2 \times 24} = \frac{720}{48} = 15 $$
Answer: \(\binom{6}{2} = 15\)
Example 3: Calculate \(\binom{7}{4}\).
Formula:
$$ \binom{7}{4} = \frac{7!}{4! \times 3!} $$
Calculate factorials:
\(7! = 5040\), \(4! = 24\), \(3! = 6\)
Calculate:
$$ \frac{5040}{24 \times 6} = \frac{5040}{144} = 35 $$
Answer: \(\binom{7}{4} = 35\)
Example 4: Calculate \(\binom{10}{0}\).
Formula:
$$ \binom{10}{0} = \frac{10!}{0! \times 10!} $$
Since \(0! = 1\),
Calculate:
$$ \frac{10!}{1 \times 10!} = 1 $$
Answer: \(\binom{10}{0} = 1\)
Example 5: Calculate \(\binom{8}{3}\).
Formula:
$$ \binom{8}{3} = \frac{8!}{3! \times 5!} $$
Calculate factorials:
\(8! = 40320\), \(3! = 6\), \(5! = 120\)
Calculate:
$$ \frac{40320}{6 \times 120} = \frac{40320}{720} = 56 $$
Answer: \(\binom{8}{3} = 56\)
Example 6: At Kenya Forestry Research Institute, calculate the number of ways to select 3 tree species from a set of 7 for a biodiversity study.
Formula:$$
\binom{7}{3} = \frac{7!}{3! \times 4!}
$$
Calculate factorials:
\(7! = 5040\), \(3! = 6\), \(4! = 24\)
Calculate:$$
\frac{5040}{6 \times 24} = \frac{5040}{144} = 35
$$
Answer: \(\binom{7}{3} = 35\)
This chapter introduced Pascal's triangle as a foundational tool for understanding the coefficients in binomial expansions. It then explained the binomial theorem, detailing how to expand expressions raised to any power using the triangle's coefficients. The chapter demonstrated how the binomial theorem can be applied to find roots of numbers through approximation techniques. It also covered the estimation of errors arising from small changes in values by using the binomial expansion. Additionally, the chapter explored the role of factorials in calculating coefficients and introduced combinations as a method to determine binomial coefficients systematically. Together, these concepts provide a comprehensive framework for applying binomial expansions in various engineering mathematics problems.
Using Pascal’s triangle, write the coefficients for the expansion of \((a + b)^4\). (1 mark)
Expand \((1 + x)^3\) using the binomial theorem and find the coefficient of \(x^2\). (2 marks)
Calculate the approximate value of \(\sqrt[3]{27.216}\) using the binomial theorem by expressing it as \((27 + 0.216)^{1/3}\). (3 marks)
Estimate the percentage error in the volume of a cylindrical chemical tank if the radius is increased by 2% and the height remains constant, using binomial expansion. (3 marks)
Find the number of combinations of selecting 4 catalysts from a set of 10 using factorial notation. (2 marks)
A chemical solution concentration is given by \((1 + 0.05)^5\). Use binomial expansion to approximate the concentration value. (3 marks)
Estimate the change in the area of a rectangular reactor when the length increases by 1.5% and the width decreases by 1%, using binomial expansions. (4 marks)
Using factorial and combinations, find how many different 3-component mixtures can be formed from 8 available chemical compounds. (3 marks)
Expand \((2 + x)^5\) up to the term containing \(x^3\) using the binomial theorem. (4 marks)
A chemical pipeline diameter is increased from 50 cm to 51 cm. Using binomial expansion, estimate the percentage increase in the cross-sectional area of the pipeline. (4 marks)
Row 4 coefficients are: 1, 4, 6, 4, 1.
Formula:
$$ (1 + x)^n = \sum_{k=0}^n \binom{n}{k} x^k $$
Coefficients for \(n=3\) from Pascal’s triangle: 1, 3, 3, 1.
Expansion:
$$ (1 + x)^3 = 1 + 3x + 3x^2 + x^3 $$
Coefficient of \(x^2\) is 3.
Express as:
$$ (27 + 0.216)^{1/3} = 27^{1/3} (1 + \frac{0.216}{27})^{1/3} $$
Calculate base cube root:
$$ 27^{1/3} = 3 $$
Calculate small fraction:
$$ x = \frac{0.216}{27} = 0.008 $$
Use binomial expansion for \((1 + x)^{1/3}\) up to second term:
$$ (1 + x)^{1/3} \approx 1 + \frac{1}{3}x = 1 + \frac{1}{3} \times 0.008 = 1 + 0.0026667 = 1.0026667 $$
Multiply:
$$ 3 \times 1.0026667 = 3.008 $$
Answer: \(\boxed{3.008}\)
Volume of cylinder:
$$ V = \pi r^2 h $$
Radius increase: \(r \to r (1 + 0.02)\)
Height constant.
Using binomial expansion for \(r^2\):
$$ (1 + 0.02)^2 = 1 + 2 \times 0.02 + (0.02)^2 = 1 + 0.04 + 0.0004 = 1.0404 $$
Percentage increase in volume:
$$ (1.0404 - 1) \times 100\% = 4.04\% $$
Answer: \(\boxed{4.04\%}\)
Formula:
$$ \binom{10}{4} = \frac{10!}{4! \times (10-4)!} = \frac{10!}{4! \times 6!} $$
Calculate factorials:
$$ 10! = 3628800, \quad 4! = 24, \quad 6! = 720 $$
Compute:
$$ \binom{10}{4} = \frac{3628800}{24 \times 720} = \frac{3628800}{17280} = 210 $$
Answer: \(\boxed{210}\)
Binomial expansion:
$$ (1 + x)^5 = 1 + 5x + 10x^2 + 10x^3 + 5x^4 + x^5 $$
Use \(x=0.05\):
Calculate terms:
$$ 5 \times 0.05 = 0.25 $$
$$ 10 \times (0.05)^2 = 10 \times 0.0025 = 0.025 $$
$$ 10 \times (0.05)^3 = 10 \times 0.000125 = 0.00125 $$
$$ 5 \times (0.05)^4 = 5 \times 0.00000625 = 0.00003125 $$
$$ (0.05)^5 = 0.0000003125 $$
Sum:
$$ 1 + 0.25 + 0.025 + 0.00125 + 0.00003125 + 0.0000003125 = 1.2762815625 $$
Answer: \(\boxed{1.2763}\) (rounded to 4 decimal places)
Area \(A = L \times W\)
Change in length:
$$ L' = L (1 + 0.015) $$
Change in width:
$$ W' = W (1 - 0.01) $$
New area:
$$ A' = L' \times W' = L \times W \times (1 + 0.015)(1 - 0.01) $$
Expand using binomial approximation:
$$ (1 + 0.015)(1 - 0.01) = 1 + 0.015 - 0.01 - 0.00015 = 1 + 0.00485 $$
Percentage change in area:
$$ 0.00485 \times 100\% = 0.485\% $$
Answer: \(\boxed{0.485\% \text{ increase}}\)
Formula:
$$ \binom{8}{3} = \frac{8!}{3! \times 5!} = \frac{40320}{6 \times 120} = \frac{40320}{720} = 56 $$
Answer: \(\boxed{56}\)
Binomial theorem:
$$ (2 + x)^5 = \sum_{k=0}^{5} \binom{5}{k} 2^{5-k} x^k $$
Calculate terms for \(k=0\) to \(3\):
$$ \binom{5}{0} 2^5 x^0 = 1 \times 32 \times 1 = 32 $$
$$ \binom{5}{1} 2^4 x = 5 \times 16 \times x = 80x $$
$$ \binom{5}{2} 2^3 x^2 = 10 \times 8 \times x^2 = 80 x^2 $$
$$ \binom{5}{3} 2^2 x^3 = 10 \times 4 \times x^3 = 40 x^3 $$
Expansion up to \(x^3\):
$$ 32 + 80x + 80x^2 + 40x^3 $$
Area of circle:
$$ A = \pi r^2 $$
Initial radius \(r = 50/2 = 25\) cm
New radius \(r' = 51/2 = 25.5\) cm
Percentage increase in radius:
$$ \frac{25.5 - 25}{25} = \frac{0.5}{25} = 0.02 = 2\% $$
Area increase using binomial expansion:
$$ (1 + 0.02)^2 = 1 + 2 \times 0.02 + (0.02)^2 = 1 + 0.04 + 0.0004 = 1.0404 $$
Percentage increase in area:
$$ (1.0404 - 1) \times 100\% = 4.04\% $$
Answer: \(\boxed{4.04\%}\)
Question 11 (Compulsory - 20 marks)
Kenya Pipeline Company is evaluating the effect of small changes in pipeline diameter on flow volume. The flow volume \(V\) varies as \(V \propto (1 + x)^4\), where \(x\) is the fractional increase in diameter. For \(x = 0.03\):
a) Use the binomial theorem to expand \((1 + x)^4\) up to the fourth term and calculate the approximate increase in flow volume. (10 marks)
b) Estimate the percentage error if only the first two terms of the expansion are used instead of all four terms. (10 marks)
Question 12 (20 marks)
A chemical engineer needs to calculate the cube root of \(1.08\) using the binomial theorem.
a) Derive the expansion formula for \(\sqrt[3]{1 + x}\) up to the third term. (10 marks)
b) Calculate the approximate value of \(\sqrt[3]{1.08}\) using the derived formula. (10 marks)
Question 13 (20 marks)
A laboratory is testing mixtures of chemicals where the number of ways to select 5 chemicals out of 12 is required.
a) Calculate \(12!\), \(5!\), and \(7!\) factorial values. (6 marks)
b) Using factorials, compute the number of combinations possible. (6 marks)
c) Explain how factorial and combinations relate to the binomial theorem expansion coefficients. (8 marks)
Question 14 (20 marks)
An industrial process involves the expression \((1 - 0.025)^7\).
a) Expand this expression using the binomial theorem up to four terms. (12 marks)
b) Calculate the approximate value of the expression using the expansion. (8 marks)
Question 11
a) Expansion:$$(1 + x)^4 = 1 + 4x + 6x^2 + 4x^3 + x^4$$
For \(x=0.03\):$$= 1 + 4(0.03) + 6(0.03)^2 + 4(0.03)^3 + (0.03)^4$$$$= 1 + 0.12 + 0.0054 + 0.000108 + 0.00000081 = 1.125509$$
Increase in flow volume ≈ 12.55%
b) Using first two terms only:$$1 + 4(0.03) = 1 + 0.12 = 1.12$$
Error = \(1.125509 - 1.12 = 0.005509\)
Percentage error = \(\frac{0.005509}{1.125509} \times 100 = 0.49\%\)
Question 12
a) Binomial expansion for \((1 + x)^{1/3}\):$$1 + \frac{1}{3}x - \frac{1}{9}x^2 + \frac{5}{81}x^3$$
b) For \(x=0.08\):$$1 + \frac{1}{3}(0.08) - \frac{1}{9}(0.08)^2 + \frac{5}{81}(0.08)^3$$$$= 1 + 0.02667 - 0.000711 + 0.000315 = 1.026274$$
Question 13
a)$$12! = 479001600$$$$5! = 120$$$$7! = 5040$$
b) Number of combinations:$$\binom{12}{5} = \frac{12!}{5! \times 7!} = \frac{479001600}{120 \times 5040} = 792$$
c) The coefficients in the binomial expansion \((a + b)^n\) are given by \(\binom{n}{r}\), which represent the number of ways to select \(r\) terms from \(n\). Factorials are used to calculate these combinations.
Question 14
a) Expansion of \((1 - 0.025)^7\):$$(1 - x)^7 = 1 - 7x + 21x^2 - 35x^3 + \ldots$$
For \(x=0.025\):$$= 1 - 7(0.025) + 21(0.025)^2 - 35(0.025)^3$$$$= 1 - 0.175 + 0.013125 - 0.000547 = 0.837578$$
b) Approximate value ≈ 0.8376
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