The coefficient of $x^n$ in expansion of $(1+x)(1-x{)}^n$ is

The expression is $(1+x)(1-x{)}^n.$ Expand $(1-x{)}^n$ using binomial theorem:$(1-x{)}^n={}^n{C}_0-{}^n{C}_{1}x+{}^n{C}_2{x}^2-...+(-1{)}^{n-1}{}^n{C}_{n-1}{x}^{n-1}+(-1{)}^n{}^n{C}_n{x}^n$.
Multiply this by $(1+x):$
$(1+x)[{}^n{C}_0-...+(-1{)}^{n-1}{}^n{C}_{n-1}{x}^{n-1}+(-1{)}^n{}^n{C}_n{x}^n]$.
To find the coefficient of $x^n,$ consider the terms that result in $x^n:$$1\cdot [(-1{)}^n{}^n{C}_n{x}^n]\Rightarrow$ Coeff $=(-1{)}^n{}^n{C}_n$.$x\cdot [(-1{)}^{n-1}{}^n{C}_{n-1}{x}^{n-1}]\Rightarrow$ Coeff $=(-1{)}^{n-1}{}^n{C}_{n-1}$.Total coefficient of $x^n=(-1{)}^n(1)+(-1{)}^{n-1}(n)$.
$=(-1{)}^n-n(-1{)}^n=(-1{)}^n(1-n)$.