Question $1$

Consider $\frac{1}{1-x}={\displaystyle \underset{n=0}{\overset{}{}}}{x}^n.$ Then $\frac{1}{(1+x{)}^2}=$

A$.{\displaystyle \underset{n=1}{\overset{}{}}}n(-1{)}^n{x}^{n-1}$

B$.-{\displaystyle \underset{n=1}{\overset{}{}}}n{x}^{n-1}$

C$.{\displaystyle \underset{n=0}{\overset{}{}}}\frac{x^{n+1}}{n+1}$

D$.$

$-{\displaystyle \underset{n=1}{\overset{}{}}}n(-1{)}^n{x}^{n-1}$