 Rivka Thorpe

2020-11-20

Use the binomial series to find the Maclaurin series for the function.
$f\left(x\right)=\frac{1}{\left(1+x{\right)}^{4}}$ pierretteA

Expert

We have to find Maclaurin series of the Function $f\left(x\right)=\frac{1}{\left(1+x{\right)}^{4}}=\left(1+x{\right)}^{-4}$ by using binomial series.
We know the binomial series
$\left(1+x{\right)}^{\alpha }=\sum _{n=0}^{\mathrm{\infty }}\left(\begin{array}{c}\alpha \\ n\end{array}\right){x}^{n}$
Where $\left(\begin{array}{c}\alpha \\ n\end{array}\right)=\frac{\alpha \left(\alpha -1\right)\left(\alpha -2\right)...\left(\alpha -n+1\right)}{n!}$
We have given function
$f\left(x\right)=\frac{1}{\left(1+x{\right)}^{4}}=\left(1+x{\right)}^{-4}$
By binomial series
$\left(1+x{\right)}^{-4}=\sum _{n=0}^{\mathrm{\infty }}\left(\begin{array}{c}-4\\ n\end{array}\right){x}^{n}$
$=\sum _{n=0}^{\mathrm{\infty }}\frac{\left(-4\right)\left(-4-1\right)\left(-4-2\right)...\left(-4-n+1\right)}{n!}{x}^{n}$
$=\sum _{n=0}^{\mathrm{\infty }}\frac{\left(-4\right)\left(-4-1\right)\left(-4-2\right)...\left(-4-\left(n-1\right)\right)}{n!}{x}^{n}$
$=\sum _{n=0}^{\mathrm{\infty }}\frac{\left(-1{\right)}^{n}4.5.6.7...\left(4+\left(n-1\right)\right)}{n!}{x}^{n}$
Therefore $f\left(x\right)=\sum _{n=0}^{\mathrm{\infty }}\frac{\left(-1{\right)}^{n}4.5.6.7...\left(4+\left(n-1\right)\right)}{n!}{x}^{n}$ Jeffrey Jordon

Expert