Chapter 8.7, Problem 22E

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 22E

Thepower series for the function $\frac{1}{{\left(1+x\right)}^4}$ is $1-4x+{\displaystyle \sum _{n=2}^{\infty }\frac{{\left(-1\right)}^{n-1}\left(n+3\right)}{3^{n}n!}}{x}^n$

And the radius of it converges to $R=1$ .

Explanation of Solution

Given:

Thegiven expression is $\frac{1}{{\left(1+x\right)}^4}$ .

Calculation:

Consider the given expression is,

  $\frac{1}{{\left(1+x\right)}^4}={\left(1+x\right)}^{-4}$

Consider the formula for the binomial expansion is,

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

Substitute $k=-4$ in the above equation as,

  $\begin{array}{c}{\left(1+x\right)}^{-4}={\displaystyle \sum _{n=0}^{\infty }\left(\begin{array}{c}-4\\ n\end{array}\right)x^n}\\ =1+\left(-4\right)x+\frac{\left(-4\right)\left(-5\right)}{2!}{x}^2+\frac{\left(-4\right)\left(-5\right)\left(-6\right)}{3!}{x}^3\mathrm{...}\\ =1-4x+\frac{\left(2+3\right)!}{2!3!}{x}^2-\frac{\left(3+3\right)!}{3!3!}{x}^3+\mathrm{...}\\ =1-4x+{\displaystyle \sum _{n=2}^{\infty }\frac{{\left(-1\right)}^{n-1}\left(n+3\right)}{3^{n}n!}}{x}^n\end{array}$

Consider the binomial series expansion ${\left(1+x\right)}^k={\displaystyle \sum _{n=0}^{\infty }\left(\begin{array}{c}-4\\ n\end{array}\right)x^n}$ observe that this series converges for $\left|x\right|<1$ thus, the radius is $R=1$ .

Thus, the power series for the function $\frac{1}{{\left(1+x\right)}^4}$ is $1-4x+{\displaystyle \sum _{n=2}^{\infty }\frac{{\left(-1\right)}^{n-1}\left(n+3\right)}{3^{n}n!}}{x}^n$

And the radius of it converges to $R=1$ .