# Use the binomial series to find the Maclaurin series for the function. F(x) = fraction {1}{(1 +...

## Question:

Use the binomial series to find the Maclaurin series for the function. {eq}\displaystyle f(x) = \frac {1}{(1 + x^5)^2} \\ \displaystyle f(x) = \sum_{n =0}^{\infty} \ \rule{3cm}{0.2mm} {/eq}

## Using the Binomial Series:

The Binomial Series provides a general formula for expanding a power of a binomial. If the power of the binomial is noninteger or negative, then the series is an infinite series. The formula is

{eq}(a + b)^r = a^r + r a^{r - 1}b + \displaystyle\frac{r(r - 1)}{2!} a^{r - 2} b^2 + \frac{r(r - 1)(r - 2)}{3!} a^{r - 3} b^3 + \ldots {/eq}

In this problem the power {eq}r {/eq} is a negative integer, so when we subtract a positive integer from r, the value continues to grow larger and larger, and remains negative.

## Answer and Explanation:

We are given the function {eq}f(x) = (1 + x^5)^{-2}, {/eq} so {eq}a = 1, \: b = x^5, {/eq} and {eq}r = -2. {/eq} Therefore the first few terms of the binomial series are

{eq}\begin{eqnarray*}(1 + x^5)^{-2} & = & 1 + (-2) (x^5) + \displaystyle\frac{(-2)(-3)}{2!} (x^5)^2 + \frac{(-2)(-3)(-4)}{3!} (x^5)^3 + \ldots \\ & = & 1 - 2x^5 + 3 x^{10} - 4 x^{15} + \ldots \end{eqnarray*} {/eq}

The general term of the series is {eq}\displaystyle\frac{(-2)(-2 - 1)(-2 - 2) \ldots (-2 - n + 1)}{n!} (x^5)^n = \frac{(-1)^n (2)(3)(4) \ldots (n +1) x^{5n}}{n!} = (-1)^n (n + 1) x^{5n}. {/eq} We can write the Maclaurin series in summation form as

{eq}f(x) = \displaystyle\sum_{n = 0}^\infty (-1)^n (n + 1) x^{5n}. {/eq}

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from Algebra II Textbook

Chapter 21 / Lesson 16