$\left ( 1+x \right )^{n}=^{n}\textrm{C}_{0}+^{n}\textrm{C}_{1}.x+^{n}\textrm{C}_{2}.x^{2}+^{n}\textrm{C}_{3}.x^{3}+.....+^{n}\textrm{C}_{n}.x^{n}$
Multiplying both sides by $x$
$x\left ( 1+x \right )^{n}=^{n}\textrm{C}_{0}.x+^{n}\textrm{C}_{1}.x^{2}+^{n}\textrm{C}_{2}.x^{3}+^{n}\textrm{C}_{3}.x^{4}+.....+^{n}\textrm{C}_{n}.x^{n+1}$
Integrating both sides w.r.t. $x$ we get
$\frac{x.(1+x)^{(n+1)}}{n+1}-\frac{(1+x)^{(n+2)}}{(n+1).(n+2)}+k=\frac{^{n}\textrm{C}_{0}.x^{2}}{2}+\frac{^{n}\textrm{C}_{1}.x^{3}}{3}+\frac{^{n}\textrm{C}_{2}.x^{4}}{4}+.....+\frac{^{n}\textrm{C}_{n}.x^{n+2}}{n+2}$ , where $k$ is arbitrary constant.
Now putting $x=0$ on both sides, $k=\frac{1}{(n+1)(n+2)}$
Again putting $x=-1$ we get,
$\frac{1}{(n+1)(n+2)}=\frac{C_{0}}{2}-\frac{C_{1}}{3}+\frac{C_{2}}{4}-......+(-1)^{n+2}\frac{C_{n}}{n+2}$
$\frac{1}{(n+1)(n+2)}=\frac{C_{0}}{2}-\frac{C_{1}}{3}+\frac{C_{2}}{4}-......+(-1)^{n}\frac{C_{n}}{n+2}$
So, Answer will be $D)$