Chapter 8.7, Problem 23E

To determine

To find: The binomial series to expand the function as the power series and state the radius of the convergence.

Answer to Problem 23E

The required binomial series is ${\displaystyle \sum _{n=0}^{\infty }\frac{{\left(-1\right)}^n\left(n+1\right)\left(n+2\right)x^n}{2^{n+4}}}$ with $R=2$ .

Explanation of Solution

Given:

The given expression is $\frac{1}{{\left(2+x\right)}^3}$ .

Calculation:

Consider the given expression is,

  $\begin{array}{c}\frac{1}{{\left(2+x\right)}^3}=\frac{1}{{\left[2\left(1+\frac{x}{2}\right)\right]}^3}\\ =\frac{1}{8}{\left(1+\frac{x}{2}\right)}^{-3}\end{array}$

Consider the formula for the binomial expansion is,

  $\begin{array}{c}{\left(1+x\right)}^k={\displaystyle \sum _{n=0}^{\infty }\left(\begin{array}{c}k\\ n\end{array}\right)x^n}\\ =1+kx+\frac{k\left(k-1\right)}{2!}{x}^2+\frac{k\left(k-1\right)\left(k-2\right)}{3!}{x}^3\mathrm{...}\end{array}$

Then, the binomial coefficient is,

  $\begin{array}{c}\left(\begin{array}{c}-3\\ n\end{array}\right)=\frac{\left(-3\right)\left(-4\right)\left(-5\right)\mathrm{.....}\left(-3-n+1\right)}{n!}\\ =\frac{\left(-3\right)\left(-4\right)\left(-5\right)\mathrm{.....}\left(-\left(n+2\right)\right)}{n!}\cdot \frac{2}{2}\\ =\frac{{\left(-1\right)}^n\left(n+1\right)\left(n+2\right)}{2}\end{array}$

Then,

  $\begin{array}{c}\frac{1}{{\left(2+x\right)}^3}=\frac{1}{8}{\displaystyle \sum _{n=0}^{\infty }\frac{{\left(-1\right)}^n\left(n+1\right)\left(n+2\right)}{2}}\frac{x^n}{2^n}\\ ={\displaystyle \sum _{n=0}^{\infty }\frac{{\left(-1\right)}^n\left(n+1\right)\left(n+2\right)x^n}{2^{n+4}}}\end{array}$

The above equation is of the interval,

  $\begin{array}{l}\frac{x}{2}<1\\ \left|x\right|<2\end{array}$

Thus, the required binomial series is ${\displaystyle \sum _{n=0}^{\infty }\frac{{\left(-1\right)}^n\left(n+1\right)\left(n+2\right)x^n}{2^{n+4}}}$ with $R=2$ .