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Chapter 2: Problem 4

Determine the expansion of \(\left(\frac{w}{4}-\frac{x}{3}\right)^{7}\).

Short Answer

Expert verified
Use the Binomial Theorem with \ a = \frac{w}{4}, b = -\frac{x}{3}, n = 7 \ and sum the terms.

Step by step solution

01

- Identify the Binomial Theorem

Recognize that the Binomial Theorem can be used to expand \[ \left( \frac{w}{4} - \frac{x}{3} \right)^7 \] which states \[ (a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]
02

- Assign Variables

In this case, assign \[ a = \frac{w}{4} \] and \[ b = -\frac{x}{3} \], with \[ n = 7 \]
03

- Write the General Term

Using the Binomial Theorem, the general term in the expansion is written as \[ \binom{7}{k} \left( \frac{w}{4} \right)^{7-k} \left( -\frac{x}{3} \right)^k \]
04

- Simplify the General Term

Simplify the general term \[ \binom{7}{k} \left( \frac{w}{4} \right)^{7-k} \left( -\frac{x}{3} \right)^k = \binom{7}{k} \left( \frac{w^{7-k}}{4^{7-k}} \right) \left( -\frac{x^k}{3^k} \right) = \binom{7}{k} \left( \frac{w^{7-k}}{4^{7-k}} \right) \left( \frac{(-1)^k x^k}{3^k} \right) = \binom{7}{k} \left( \frac{w^{7-k} (-1)^k x^k}{4^{7-k} 3^k} \right) \]
05

- Write Out Terms

Expand by writing out each term for \ k = 0 \ to \ 7 \. Example, for \ k = 0: \[ \binom{7}{0} \left( \frac{w^{7-0}}{4^{7-0}} \right) \left( -\frac{x^0}{3^0} \right) = 1 \left( \frac{w^7}{4^7} \right) \left( 1 \right) = \frac{w^7}{16384} \]. Continue this process for \ k = 1 \ to \ 7.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Binomial Theorem
Understanding the binomial theorem is essential for expanding expressions like \ \(\left( \frac{w}{4} - \frac{x}{3} \right)^{7}\). The binomial theorem provides a way to expand expressions of the form \ (a+b)^n \ into a sum involving terms of the form \ \binom{n}{k} a^{n-k} b^k \. Here, each term in the expansion is determined by the binomial coefficient \ \binom{n}{k} \, which is calculated using combinations: \ \binom{n}{k} = \frac{n!}{k!(n-k)!} \. This coefficient determines how many ways you can choose \ k \ items from \ n \ items.
General Term
The general term in a binomial expansion is crucial for writing any term in the series. For \ \(\left( \frac{w}{4} - \frac{x}{3} \right)^{7}\), identifying \ a = \frac{w}{4} \, \ b = -\frac{x}{3} \, and \ n = 7 \ is important. Using the binomial theorem formula \ \binom{7}{k} \, the general term for this expression can be written as: \ \[ \binom{7}{k} \left( \frac{w}{4} \right)^{7-k} \left( -\frac{x}{3} \right)^k. \ \ \] By evaluating the general term, \ k \ can take any integer value from \ 0 \ to \ 7\.
Polynomial Expansion
The expansion of \ \(\left( \frac{w}{4} - \frac{x}{3} \right)^{7}\) involves simplifying and writing out each term. Start with computing the binomial coefficients and substituting values. For example, the term when \ k = 0 \ is: \ \[ \binom{7}{0} \left( \frac{w^{7-0}}{4^{7-0}} \right) \left( -\frac{x^0}{3^0} \right) = 1 \left( \frac{w^7}{4^7} \right) \left( 1 \right) = \frac{w^7}{16384} \. \ \ \] Repeat this process for \ k = 1 \ to \ 7\, simplifying each term individually. This gives the entire polynomial expansion as a series of terms, each determined by the respective values of \ k \.

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Most popular questions from this chapter

[mechanics] The position, \(s\), of a particle along a line is given by $$ s= \begin{cases}t^{2}+1 & 0 \leq t<2 \\ 2 t+1 & 2 \leq t<10\end{cases} $$ Sketch the graph of \(s\) against \(t\) for \(0 \leq t<10\)

[mechanics] The height \(h\) of a projectile fired from the ground with respect to time \(t\) is given by $$ h=12 t-t^{2} $$Sketch this curve. Find the maximum height reached by the projectile.

[electrical principles] The voltage, \(V\), across a variable load resistor of resistance \(R_{\mathrm{L}}\), is given by $$ V=\frac{E R_{\mathrm{L}}}{R+R_{\mathrm{L}}} $$ where \(E\) is the source e.m.f. and \(R\) is the source resistance. Plot the graphs (on different axes) of \(V\) versus \(R_{\mathrm{L}}\) for the corresponding values: a \(E=60\) volts, \(R=10 \Omega\) for \(0 \leq R_{\mathrm{L}} \leq 20\) b \(E=15\) volts, \(R=3 \times 10^{3} \Omega\) for \(0 \leq R_{\mathrm{L}} \leq 6 \times 10^{3}\) c \(E=10\) volts, \(R=15 \times 10^{3} \Omega\) for \(0 \leq R_{\mathrm{L}} \leq 30 \times 10^{3}\) From each of your graphs, determine the value of \(V\) at \(R_{\mathrm{L}}=R\). Do you notice any relationship between \(V\) and \(E\) at \(R_{\mathrm{L}}=R\) ? Show algebraically that if \(R_{\mathrm{L}}=R\) then \(V=\frac{E}{2}\).

across a capacitor is given by $$ V= \begin{cases}2 t & 0 \leq t<1 \\ 4-2 t & t \geq 1\end{cases} $$ Sketch the graph of \(V\) against \(t\) for \(0 \leq t \leq 2\) [Hint: The graph of \(V\) is given in two parts. You sketch \(V=2 t\) for \(0 \leq t<1\) and then sketch \(V=4-2 t\) for \(t \geq 1 .]\)

of a circuit is defined by: $$ V=t^{2}-5 t+6 \quad(t \geq 0) $$ Sketch the graph of \(V\) against \(t\), indicating the minimum value of \(V\).
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