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Chapter 14: Problem 34

Write the first three terms in each binomial expansion, expressing the result in simplified form. $$(x-2 y)^{9}$$

Short Answer

Expert verified
The first three terms in the expansion of \((x-2y)^9\) are \(x^9, -18x^8y, 144x^7y^2\).

Step by step solution

01

Recall the Binomial Theorem

The Binomial Theorem allows us to expand a binomial raised to any positive integer power, \(n\). The general formula for the expansion is \((a+b)^n = \sum_{k=0}^{n} {n \choose k} a^{n-k} b^k\), where \({n \choose k}\) are the binomial coefficients or combinations.
02

Apply the Binomial Theorem to the given binomial

Let's apply this theorem now to the binomial \((x - 2y)^9\). Here \(a=x\), \(b=-2y\), and \(n=9\). Keep in mind that we only need to calculate first three terms. \n \n - The first term is: \({9 \choose 0} (x)^{9-0} (-2y)^0 = 1 * x^9 * 1 = x^9\). \n - The second term is: \({9 \choose 1} (x)^{9-1} (-2y)^1 = 9 * x^8 * -2y = -18x^8y\). \n - The third term is: \({9 \choose 2} (x)^{9-2} (-2y)^2 = 36 * x^7 * 4y^2 = 144x^7y^2 \).
03

Final Simplified Result

Combining these terms, the first three terms of the binomial expansion are \(x^9, -18x^8y, 144x^7y^2\).

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Most popular questions from this chapter

Use a graphing utility to graph the function. Determine the horizontal asymptote for the graph of \(f\) and discuss its relationship to the sum of the given series. Function $$f(x)=\frac{2\left[1-\left(\frac{1}{3}\right)^{x}\right]}{1-\frac{1}{3}}$$ Series$$\begin{array}{l}2+2\left(\frac{1}{3}\right)+2\left(\frac{1}{3}\right)^{2} \\\\+2\left(\frac{1}{3}\right)^{3}+\cdots\end{array}$$

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