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Chapter 9: Problem 38

Expanding an Expression In Exercises \(19-40,\) use the Binomial Theorem to expand and simplify the expression. $$\left(\frac{2}{x}-3 y\right)^{5}$$

Short Answer

Expert verified
\[32 x^{-5} - 240 y x^{-4} + 720 y^{2} x^{-3} - 1080 y^{3} x^{-2} + 810 y^{4} x^{-1} - 243 y^{5}\]

Step by step solution

01

Identify the components

Identify a, b, and n where the expression is (a+b)^n. For our expression \(\left(\frac{2}{x}-3 y\right)^{5}\), a = \(\frac{2}{x}\), b = -3y (note the negative sign) and n = 5.
02

Apply the Binomial Theorem

Based on the Binomial theorem, the expanded form of the given expression would be the sum of terms which can be calculated as \(\binom{n}{k}\) multiplied by \(a^{n-k}\) and \(b^k\) respectively. Here, \(k\) takes on values starting from 0 up to \(n\). Now substituting our values of a, b and n, we have: \[\binom{5}{0} (\frac{2}{x})^{5} (-3y)^{0} + \binom{5}{1} (\frac{2}{x})^{4} (-3y)^{1} + \binom{5}{2} (\frac{2}{x})^{3} (-3y)^{2} + \binom{5}{3} (\frac{2}{x})^{2} (-3y)^{3} + \binom{5}{4} (\frac{2}{x})^{1} (-3y)^{4} + \binom{5}{5} (\frac{2}{x})^{0} (-3y)^{5}\]
03

Perform the Computation

Now, every term can be computed respectively. The binomial coefficients, \(\binom{5}{k}\), can be calculated as 5 choose k, and \(-3y\)^k will introduce alternating signs due to the negative. After much calculations, we get: \[32 x^{-5} - 240 y x^{-4} + 720 y^{2} x^{-3} - 1080 y^{3} x^{-2} + 810 y^{4} x^{-1} - 243 y^{5}\]

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Most popular questions from this chapter

Proving a Property In Exercises \(31-40,\) use mathematical induction to prove the property for all positive integers \(n .\) $$\text{ A }{\text factor}\text{ of }\left(2^{2 n-1}+3^{2 n-1}\right) \text { is } 5 $$

Finding a Linear or Quadratic Model In Exercises \(55-60\) , decide whether the sequence can be represented perfectly by a linear or a quadratic model. If so, then find the model. $$5,13,21,29,37,45, \ldots$$

Approximation In Exercises \(79-82,\) use the Binomial Theorem to approximate the quantity accurate to three decimal places. For example, in Exercise \(79,\) use the expansion \((1.02)^{8}=(1+0.02)^{8}\) $$=1+8(0.02)+28(0.02)^{2}+\cdots+(0.02)^{8}$$ $$(1.98)^{9}$$

Expanding a Complex Number In Exercises \(73-78\) , use the Binomial Theorem to expand the complex number. Simplify your result. $$(1+i)^{4}$$

Approximation In Exercises \(79-82,\) use the Binomial Theorem to approximate the quantity accurate to three decimal places. For example, in Exercise \(79,\) use the expansion \((1.02)^{8}=(1+0.02)^{8}\) $$=1+8(0.02)+28(0.02)^{2}+\cdots+(0.02)^{8}$$ $$(1.02)^{8}$$
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