Question

In Example 1 we used Theorem 8.11 to find the Maclaurin series for $\frac{1}{(1-x{)}^2}.$Explain how Theorem 8.11 could be used to find the Maclaurin series for $\frac{1}{(1-x{)}^2}$, where $k$is a positive integer greater than $2$.

Short answer

Since $\frac{d^{k-1}}{d{x}^{k-1}}\left(\frac{1}{1-x}\right)=\frac{(k-1)\text{!}}{(1-x{)}^k},$so to find the Maclaurin series for$\frac{1}{(1-x{)}^k},$we take the$(k-1{)}^n$derivative of the series $\frac{1}{1-x}$term by term and divide each term by $k!$

Step-by-step solution

Step :1 Given Information

Given function :$\frac{1}{(1-x{)}^2}$

Explaining how Theorem 8.11 could be used to find the Maclaurin series 

The Maclaurin series for $\frac{1}{1-x}$is $\boxed{\sum _{k=0}^{\infty }{x}^k}$

So, the Maclaurin series for $\frac{1}{(1-x{)}^2}$can be found by simply differentiating the Maclaurin series for $\frac{1}{1-x}$.

This is because $\frac{d}{dx}\left(\frac{1}{1-x}\right)=\frac{1}{(1-x{)}^2}$

Similarly, since $\frac{d^{k-1}}{d{x}^{k-1}}\left(\frac{1}{1-x}\right)=\frac{(k-1)\text{!}}{(1-x{)}^k}$, so to find the Maclaurin series for $\frac{1}{(1-x{)}^k}$, we could take the ${(k-1)}^n$derivative of the series $\frac{1}{1-x}$term by term and divide each term by $k!$