# Use the binomial series to expand the function as a power series. 5 (6 + x)3 Use the binomial...

## Question:

Use the binomial series to expand the function as a power series. {eq}\dfrac 5{(6 + x)^3} {/eq}.

State the radius of convergence, R.

## Binomial Series Representation:

To get the expression of power series representation for the function {eq}\; \dfrac {1}{(6 + x)^{3}} \; {/eq}, we will use the Binomial series representation for the negative powers. First of all, we will form the series for the function {eq}\; \dfrac {1}{\Biggr[1 + \biggr( \dfrac {x}{6} \biggr) \Biggr]^{3}} \; {/eq} then we will go for the main function.

{eq}\displaystyle (1 + a)^{k} = 1 + ka + \dfrac {k (k-1)}{2!} \; a^{2} + \dfrac {k (k-1) (k -2)}{3!} \; a^{3} + \cdots {/eq}

The interval of convergence is given as: {eq}\; \; \Longrightarrow \; |a| < 1 {/eq}

## Answer and Explanation:

{eq}\\ {/eq}

{eq}\displaystyle f(x) = \dfrac {5}{(6 + x)^{3}} {/eq}

{eq}\displaystyle \dfrac {1}{(6 + x)^{3}} = \dfrac {1}{6^{3}} \; \Biggr[1 + \biggr( \dfrac {x}{6} \biggr) \Biggr]^{-3} {/eq}

We know the standard Binomial series representation for the negative powers:

{eq}\displaystyle (1 + a)^{k} = 1 + ka + \dfrac {k (k-1)}{2!} \; a^{2} + \dfrac {k (k-1) (k-2)}{3!} \; a^{3} + \cdots {/eq}

The interval of convergence is given as: {eq}\; \; \Longrightarrow \; |a| < 1 {/eq}

{eq}\displaystyle (1 + a)^{-3} = 1 - 3a + \dfrac {(-3) \; (-3 -1)}{2!} \; a^{2} + \dfrac {(-3) \; (-3-1) \; (-3-2) }{3!} \; a^{3} + \cdots {/eq}

{eq}\displaystyle (1 + a)^{-3} = 1 - 3a + 6a^{2} - 10a^{3} + \cdots {/eq}

Now replace the value of {eq}\; a = \dfrac {x}{6} \; {/eq} in the above expression:

{eq}\displaystyle \Biggr[1 + \biggr( \dfrac {x}{6} \biggr) \Biggr]^{-3} = 1 - \dfrac {x}{2} + \dfrac {x^{2}}{6} - \dfrac {5x^{3}}{108} + \cdots {/eq}

The interval of convergence is given as: {eq}\; \; |\dfrac {x}{6}| < 1 \; \; \; \Longrightarrow \; \; |x| < 6 {/eq}

Finally, the power series representation is given as:

{eq}\displaystyle \Longrightarrow \boxed {f(x) = \dfrac {1}{(6 + x)^{3}} = \dfrac {1}{6^{3}} \; \Biggr[ 1 - \dfrac {x}{2} + \dfrac {x^{2}}{6} - \dfrac {5x^{3}}{108} + \cdots \Biggr]} {/eq}

The interval of convergence is given as: {eq}\; \; \Longrightarrow \; \; |x| < 6 {/eq}

The radius of convergence is given as: {eq}\; \; \Longrightarrow \; \; \text {R} = 6 {/eq}

#### 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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