The value of $\frac{C_0}{1.3}-\frac{C_1}{2.3}+\frac{C_2}{3.3}-\frac{C_3}{4.3}+.....{\left(-1\right)}^n\frac{C_n}{(n+1).3}$ is

Consider the expansion ${\left(1-x\right)}^n=C_0-C_{1}x+C_2{x}^2-...{\left(-1\right)}^n{x}^n$,

Now, integrating the expansion with respect to $x$ from $0$ to $1$ gives,

$$\begin{gathered} \frac{1}{n+1}=\frac{C_0}{1}-\frac{C_1}{2}+\frac{C_2}{2}-...{\left(-1\right)}^n\frac{C_n}{n+1} \\ ⟹\frac{1}{3\left(n+1\right)}=\frac{C_0}{1.3}-\frac{C_1}{2.3}+\frac{C_2}{3.3}-...{\left(-1\right)}^n\frac{C_n}{\left(n+1\right).3} \end{gathered}$$