Question

In Problems 11-16 use an appropriate series in (2) to find the Maclaurin series of the given function. Write your answer in summation notation.

$\frac{\mathbf{1}}{\mathbf{2}\mathbf{+}\mathbf{x}}$

Short answer

$\frac{1}{2+x}=\sum _{n=0}^{\infty }\frac{{(-1)}^n{x}^n}{2^{n+1}}$

Step-by-step solution

Definition

Maclaurin seriesare a type of series expansion in which all terms are nonnegative integer powers of the variable.

Rewritten

Consider the following function $f\left(x\right)=\frac{1}{2+x}$,the function get

$$\begin{gathered} f\left(x\right)=\frac{1}{2+x} \\ =\frac{1}{2\left(1+\frac{x}{2}\right)} \\ =\frac{1}{2\left(1-\left(-\frac{x}{2}\right)\right)} \\ =\frac{1}{2}\frac{1}{1-\left(-\frac{x}{2}\right)} \end{gathered}$$

Thus $f\left(x\right)=\frac{1}{2}\frac{1}{1-\left(-\frac{x}{2}\right)}\dots \dots$(1)

Find Maclaurin series

We know that Maclaurin series expansion of $\frac{1}{1-x}$.

$\frac{1}{1-x}=1+x+x^2+x^3+\cdots$

Replace $x$with$-\frac{x}{2}$ in the above one,

$\frac{1}{1-\left(-\frac{x}{2}\right)}=1+\left(-\frac{x}{2}\right)+{\left(-\frac{x}{2}\right)}^2+{\left(-\frac{x}{2}\right)}^3+\cdots \dots \dots \left(2\right)$

Substitute value

From (1) and (2), get

$$\begin{gathered} f\left(x\right)=\frac{1}{2}\frac{1}{1-\left(-\frac{x}{2}\right)} \\ =\frac{1}{2}\left[1+\left(-\frac{x}{2}\right)+{\left(-\frac{x}{2}\right)}^2+{\left(-\frac{x}{2}\right)}^3+\cdots \right] \\ =\frac{1}{2}\sum _{n=0}^{\infty }{\left(-\frac{x}{2}\right)}^n \\ =\frac{1}{2}\sum _{n=0}^{\infty }\frac{{(-1)}^n{x}^n}{2^n} \\ =\sum _{n=0}^{\infty }\frac{{(-1)}^n{x}^n}{2^{n+1}} \end{gathered}$$

Hence,$\frac{1}{2+x}=\sum _{n=0}^{\infty }\frac{{(-1)}^n{x}^n}{2^{n+1}}$