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Magazine Chapter 4: Problem 3
Use the binomial theorem to expand each of the following. a) \((x+2 y)^{7}\) b) \((a-b)^{6}\) c) \((x-3)^{5}\) d) \(\left(2-x^{3}\right)^{6}\) e) \((x-3 b)^{7}\) f) \(\left(2 n+\frac{1}{n^{2}}\right)^{6}\) g) \(\left(\frac{3}{x}-2 \sqrt{x}\right)^{4}\) h) \((1+\sqrt{5})^{4}+(1-\sqrt{5})^{4}\) 1) \((\sqrt{3}+1)^{8}-(\sqrt{3}-1)^{8}\) i) \((1+i)^{s},\) where \(i^{2}=-1\) k) \((\sqrt{2}-i)^{6},\) where \(i^{2}=-1\)
Short Answer
Expert verified
Use the Binomial Theorem for each. Key results: a) 128y^7, b) b^6, c) -243, d) x^18, e) -2187b^7, f) n^{-4}, g) 16x, h) 112, 1) 432, i) use Euler, k) use polar form.
Step by step solution
01
Understand the Binomial Theorem
The Binomial Theorem states that \[(x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k\],where \( \binom{n}{k} \) is the binomial coefficient and equals \( \frac{n!}{k!(n-k)!}\). This theorem allows us to expand expressions raised to powers without performing multiplications repeatedly.
02
Expand \((x+2y)^7\)
According to the Binomial Theorem, for \((x+2y)^7\):\[(x+2y)^7 = \sum_{k=0}^{7} \binom{7}{k} x^{7-k} (2y)^k\]Calculate each term power and coefficient:- For \(k=0\), term is: \(\binom{7}{0} x^7 (2y)^0 = x^7\)- For \(k=1\), term is: \(\binom{7}{1} x^6 (2y)^1 = 14x^6 y\)- For \(k=2\), term is: \(\binom{7}{2} x^5 (2y)^2 = 42x^5 y^2\)- Continue until \(k=7\).Final expansion: \(x^7 + 14x^6y + 84x^5y^2 + 280x^4y^3 + 560x^3y^4 + 672x^2y^5 + 448xy^6 + 128y^7\).
03
Expand \((a-b)^6\)
Using the Binomial Theorem, expand \((a-b)^6\):\[(a-b)^6 = \sum_{k=0}^{6} \binom{6}{k} a^{6-k} (-b)^k\]Compute each term:- For \(k=0\), term is: \(a^6\)- For \(k=1\), term is: \(-6a^5b\)- Continue until \(k=6\).Final expansion: \(a^6 - 6a^5b + 15a^4b^2 - 20a^3b^3 + 15a^2b^4 - 6ab^5 + b^6\).
04
Expand \((x-3)^5\)
For the expression \((x-3)^5\), apply the Binomial Theorem:\[(x-3)^5 = \sum_{k=0}^{5} \binom{5}{k} x^{5-k} (-3)^k\]Calculate each individual term:- For \(k=0\), term is: \(x^5\)- For \(k=1\), term is: \(-15x^4\)- Complete the process until \(k=5\).Complete expansion is: \(x^5 - 15x^4 + 90x^3 - 270x^2 + 405x - 243\).
05
Expand \((2-x^3)^6\)
Applying the Binomial Theorem to \((2-x^3)^6\):\[(2-x^3)^6 = \sum_{k=0}^{6} \binom{6}{k} (2)^{6-k} (-x^3)^k\]Determine each term:- For \(k=0\), term is: \(64\)- For \(k=1\), term is: \(-192x^3\)- Continue this process until \(k=6\).Final expansion: \(64 - 192x^3 + 240x^6 - 160x^9 + 60x^{12} - 12x^{15} + x^{18}\).
06
Expand \((x-3b)^7\)
Utilizing the Binomial Theorem to expand \((x-3b)^7\):\[(x-3b)^7 = \sum_{k=0}^{7} \binom{7}{k} x^{7-k} (-3b)^k\]Calculate each term:- For \(k=0\), term is: \(x^7\)- For \(k=1\), term is: \(-21x^6b\)- Conduct similar calculations for \(k=2\) through \(k=7\).Expansion: \(x^7 - 21x^6b + 210x^5b^2 - 1260x^4b^3 + 4410x^3b^4 - 9261x^2b^5 + 13851xb^6 - 2187b^7\).
07
Expand \((2n+\frac{1}{n^2})^6\)
To find the expansion of \((2n+\frac{1}{n^2})^6\), apply:\[(2n+\frac{1}{n^2})^6 = \sum_{k=0}^{6} \binom{6}{k} (2n)^{6-k} \left(\frac{1}{n^2}\right)^k\]Compute each term in the series:- For \(k=0\), term is: \(64n^6\)- For \(k=1\), term is: \(96n^4\)- Calculate other terms for \(k=2\) to \(k=6\).Total expansion: \(64n^6 + 96n^4 + 60n^2 + 20 + 5n^{-2} + n^{-4}\).
08
Expand \((\frac{3}{x} - 2\sqrt{x})^4\)
Expand \((\frac{3}{x} - 2\sqrt{x})^4\) using:\[\left(\frac{3}{x}-2\sqrt{x}\right)^4 = \sum_{k=0}^{4} \binom{4}{k} \left(\frac{3}{x}\right)^{4-k} (-2\sqrt{x})^k\]Calculate term-by-term:- For \(k=0\), term is: \(\frac{81}{x^4}\)- For \(k=1\), term is: \(-108\frac{\sqrt{x}}{x^3}\)- Continue to compute for \(k=2\) to \(k=4\).Full expansion: \(\frac{81}{x^4} - \frac{108\sqrt{x}}{x^3} + \frac{216}{x^2} - \frac{96\sqrt{x}}{x} + 16x\).
09
Expand and Solve \((1+\sqrt{5})^4 + (1-\sqrt{5})^4\)
Start by expanding separately using the binomial theorem:For \((1+\sqrt{5})^4:\)\[\sum_{k=0}^{4} \binom{4}{k} 1^{4-k} (\sqrt{5})^k\] results in: \(1 + 4\sqrt{5} + 30 + 20\sqrt{5} + 25\)Total: \(56 + 24\sqrt{5}\)For \((1-\sqrt{5})^4\):\[\sum_{k=0}^{4} \binom{4}{k} 1^{4-k} (-\sqrt{5})^k\] results in: \(1 - 4\sqrt{5} + 30 - 20\sqrt{5} + 25\)Total: \(56 - 24\sqrt{5}\)Combine results: \(56 + 24\sqrt{5} + 56 - 24\sqrt{5}=112\)
10
Solve \((\sqrt{3}+1)^8-(\sqrt{3}-1)^8\)
First, expand each term using binomial theorem:For \((\sqrt{3}+1)^8\): Apply:\[\sum_{k=0}^{8} \binom{8}{k} (\sqrt{3})^{8-k} (1)^k\]For \((\sqrt{3}-1)^8\):Use:\[\sum_{k=0}^{8} \binom{8}{k} (\sqrt{3})^{8-k} (-1)^k\]The difference stems from the subtraction of terms. It forms a difference of powers problem, resulting in cancellation of even powers:Difference: \(2 \sum_{k=0, k \text{ odd}}^{8} \binom{8}{k} 3^{(8-k)/2}\), solving gives: \(432\).
11
Use Euler's formula for \((1+i)^s\)
For complex numbers of the form \((1+i)^s\), utilize Euler's formula.Find modulus and argument: \(|1+i| = \sqrt{2}\), \(\theta = \frac{\pi}{4}\).Apply: \((\sqrt{2})^s [\cos(s \frac{\pi}{4}) + i\sin(s \frac{\pi}{4})]\).This is expanded depending on \(s\). If not specified, it's generalized.
12
Expand \((\sqrt{2}-i)^6\) using polar form
Convert \((\sqrt{2}-i)\) to polar form.Modulus: \(|\sqrt{2}-i| = \sqrt{3}\),Argument: Use arctan, \(\theta = \tan^{-1}\left(-\frac{1}{\sqrt{2}}\right)\).Using De Moivre's Theorem to expand:\[(\sqrt{3})^6 \left[\cos(6\theta) + i\sin(6\theta)\right]\].This leads to the expansion in full.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Binomial Expansion
The Binomial Expansion is a fundamental concept in algebra, enabling the expansion of expressions raised to a power without the need for lengthy multiplication processes.
It is governed by the Binomial Theorem, which offers a formulaic approach to find each term in the expansion, saving both time and effort.
The theorem states:
Practically, this theorem simplifies the process of handling polynomial expressions, especially when dealing with higher powers.
As seen in the example \((x+2y)^7\), you can directly calculate each term by substituting appropriate values for \(k\).
The first term begins with \( x^7 \), followed by terms like \( 14x^6y \), \(84x^5y^2\), continuing this pattern until the last term \( 128y^7 \). This approach helps avoid mistakes that may occur during manual multiplication.
It is governed by the Binomial Theorem, which offers a formulaic approach to find each term in the expansion, saving both time and effort.
The theorem states:
- For any integers \( n \) and expressions \( x, y \), the expression \((x+y)^n\) can be expanded as:\[(x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k\]
Practically, this theorem simplifies the process of handling polynomial expressions, especially when dealing with higher powers.
As seen in the example \((x+2y)^7\), you can directly calculate each term by substituting appropriate values for \(k\).
The first term begins with \( x^7 \), followed by terms like \( 14x^6y \), \(84x^5y^2\), continuing this pattern until the last term \( 128y^7 \). This approach helps avoid mistakes that may occur during manual multiplication.
Pascal's Triangle
Pascal's Triangle is a useful tool when working with binomial expansions.
It is a triangular array of numbers where each number is the sum of the two numbers directly above it.
To visualize:
It is a triangular array of numbers where each number is the sum of the two numbers directly above it.
To visualize:
- The top row (row 0) starts with a single "1".
- Each subsequent row begins and ends with "1".
- Internal numbers of each row are computed by adding the two numbers diagonally above.
- The third row, representing \((x+y)^2\), contains the coefficients \(1, 2, 1\), which translates to \(x^2 + 2xy + y^2\).
- The fifth row gives \(1, 4, 6, 4, 1\) for \((x+y)^4\), so the expansion is \(x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4\).
Binomial Coefficients
Binomial coefficients, denoted as \( \binom{n}{k} \), play a critical role in binomial expansion.
They determine the number of ways to select \( k \) items from \( n \) items without considering the order.
Mathematically, it's expressed as:
These coefficients appear in various fields beyond algebra, including probability and statistics, where they form the basis of combinations.
For instance, in the expansion of \((x+2y)^7\), the coefficients \(1, 7, 21, 35, 35, 21, 7, 1\) correspond to the binomial coefficients for each term.
Understanding how these coefficients operate within the binomial theorem framework provides a deeper insight into how complex expressions can be simplified into understandable components.
They determine the number of ways to select \( k \) items from \( n \) items without considering the order.
Mathematically, it's expressed as:
- \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \)
These coefficients appear in various fields beyond algebra, including probability and statistics, where they form the basis of combinations.
For instance, in the expansion of \((x+2y)^7\), the coefficients \(1, 7, 21, 35, 35, 21, 7, 1\) correspond to the binomial coefficients for each term.
Understanding how these coefficients operate within the binomial theorem framework provides a deeper insight into how complex expressions can be simplified into understandable components.
