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Chapter 7: Problem 561

Co-efficient of \(\mathrm{x}^{-3}\) in expansion of \([\mathrm{x}-(\mathrm{a} / \mathrm{x})]^{11}\) is \(\ldots \ldots\) (a) \(-792 \mathrm{a}^{5}\) (b) \(-923 \mathrm{a}^{7}\) (b) \(-792 \mathrm{a}^{6}\) (d) \(-330 \mathrm{a}^{7}\)

Short Answer

Expert verified
The coefficient of \(\mathrm{x}^{-3}\) in the expansion of \([\mathrm{x}-(\mathrm{a} / \mathrm{x})]^{11}\) is \(-330 \mathrm{a}^{7}\).

Step by step solution

01

Identify the variables to use in the binomial theorem formula

In our case, \(\mathrm{a} = \mathrm{x}\) and \(\mathrm{b} = -(\mathrm{a} / \mathrm{x})\). We will also use n=11 in the formula.
02

Find the term with exponent -3

To find the term with exponent -3, we need to find the value of k such that: \(11 - k - k = -3\) Solving for k, we get: \(k = 7\)
03

Use the binomial theorem formula to determine the coefficient

Now that we have a value for k, we can find the term in the expansion with exponent -3: \(\binom{11}{7} (\mathrm{x})^{11-7} (-(\mathrm{a} / \mathrm{x}))^{7}\) Simplifying this expression: \(\binom{11}{7} (\mathrm{x})^4 (-(\mathrm{a} / \mathrm{x}))^{7}\) \(\binom{11}{7} (\mathrm{x})^4 (\mathrm{a}^{7}) (\mathrm{x}^{-7})\) \(\binom{11}{7} (\mathrm{a}^{7}) (\mathrm{x}^{-3})\)
04

Find the coefficient of the term

Now that we have the term of interest, we can extract its coefficient: \(- \binom{11}{7} (\mathrm{a}^{7})\)
05

Compute the value of \(\binom{11}{7}\)

\(\binom{11}{7} = \dfrac{11!}{7! (11 - 7)!} = \dfrac{11!}{7! 4!} = 330\)
06

Combine the coefficient and x's exponent

Now that we have the coefficient's value, we can write the complete term in the expansion: \(-330 (\mathrm{a}^{7})\mathrm{x}^{-3}\) Therefore, the coefficient of \(\mathrm{x}^{-3}\) in the expansion of \([\mathrm{x}-(\mathrm{a} / \mathrm{x})]^{11}\) is \(\boxed{-330 \mathrm{a}^{7}}\). The answer is (d).

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