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Determine the middle term in the expansion of (\frac{1}{x + \sqrt{x}})^4.

Question:

Determine the middle term in the expansion of {eq}\displaystyle\;\left(\frac{1}{x + \sqrt{x}}\right)^4 {/eq}.

Binomial Expansion:

In {eq}\left ( a+b \right )^n {/eq} where n is even

middle term is given by {eq}\left ( \frac{n}{2}+1 \right )^{th} {/eq} term.

Also, rth term is given by {eq}n_{r}^{C}\textrm{}\,(a)^{n-r}b^r {/eq}

Answer and Explanation:

In {eq}(\frac{1}{x + \sqrt{x}})^4 {/eq}, n = 4 is even

So, middle term is the {eq}\left ( \frac{4}{2}+1 \right )^{th}=3^{rd}\,\,term {/eq}

Also, {eq}\displaystyle\;\left(\frac{1}{x + \sqrt{x}}\right)^4 =\left ( \frac{x-\sqrt{x}}{\left (x + \sqrt{x} \right )\left ( x - \sqrt{x} \right )} \right )=\frac{x-\sqrt{x}}{x^2-x} {/eq}

{eq}T_3=4_{2}^{C}\,\left ( \frac{x}{x^2-x} \right )^{4-2}\left ( \frac{\sqrt{x}}{x^2-x} \right )^r\\ =\frac{6}{(x-1)^2}\frac{x}{x^2(x-1)^2}\\ =\frac{6}{x(x-1)^4} {/eq}


Learn more about this topic:

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How to Use the Binomial Theorem to Expand a Binomial

from Algebra II Textbook

Chapter 21 / Lesson 16
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