# a. Expand 5 4 sqrt 1 + x as power series. 1. 5 - 5 4 + sum infinity n = 2 5 ( - 1 ) n...

## Question:

(a) Expand

{eq}\frac{5}{\sqrt[4]{1+x}} {/eq} as power series.

(a1)

{eq}5-\frac{5}{4}x +\sum_{n=2}^{\infty }5(-1)^n\frac{1.5.9...(4n-3)}{4^n.n!}x^n {/eq}

(a2)

{eq}5-\frac{5}{4}x +\sum_{n=2}^{\infty }5(-1)^n\frac{(2n+1)!}{4^n.n!}x^n {/eq}

(a3)

{eq}5-\frac{5}{4}x +\sum_{n=2}^{\infty }5(-1)^n\frac{1.4.7....(3n-2)}{4^n.n!}x^n {/eq}

(a4)

{eq}5-\frac{5}{4}x +\sum_{n=2}^{\infty }5\frac{(2n+1)!}{4^n.n!}x^n {/eq}

(a5)

{eq}\sum_{n=0}^{\infty }5(-1)^{n+1}\frac{1.5.9....(4n+1)}{4^n.n!}x^n {/eq}

(b) Use a Part (a) to estimate

{eq}J= \frac{5}{\sqrt[4]{1.1}} {/eq} correct to three decimal places.

## Binomial Series; Approximations:

We'll first find the Maclaurin series of {eq}\frac{1}{\sqrt[4]{1+x}}=(1 + x)^{-\frac{1}{4}}, {/eq} recalling the binomial expansion for any non-zero real number {eq}p {/eq}:

{eq}\begin{align*} (1+x)^p&=\sum_{n=0}^{\infty}\binom{p}{n}x^n\\ &=1+\sum_{n=1}^{\infty}\frac{p(p-1)(p-2)\cdots (p-n+1)}{n!}x^n\\ &=1+px+\frac{p(p-1)}{2}x^2 +\frac{p(p-1)(p-2)}{3!}x^{3}+\\ &\qquad\cdots+\frac{p(p-1)(p-2)\cdots (p-n+1)}{n!}x^n+\cdots\\ &\text{and converges for }|x|<1. \end{align*} {/eq}

We'll multiply by {eq}5 {/eq} to get the Maclaurin expansion for {eq}\frac{5}{\sqrt[4]{1+x}} {/eq}.

To estimate {eq}\frac{5}{\sqrt[4]{1.1}} {/eq} correct to three decimal places, we'll use the Alternating Series Theorem to select a partial sum with an error smaller than {eq}10^{-4} {/eq} (to get three correct decimal places).

## Answer and Explanation:

For Part (a): The valid option is (a1) as we'll explain below.

First we use the binomial expansion:

{eq}\begin{align*} (1+x)^p&=\sum_{n=0}^{\infty}...

See full answer below.

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#### Learn more about this topic:

How to Use the Binomial Theorem to Expand a Binomial

from Algebra II Textbook

Chapter 21 / Lesson 16
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