If $A$ is idemponent matrix, then find the value of  ${(A+I)}^n$$, \forall n\in N$ Where $I$ is the identity matrix having the same order of $A$.

$\therefore A$ is idemponent matrix

$\therefore A^2=A$,

similarly $A=A^2=A^3=A^4=...=A^n ...\left(i\right)$

Now, ${(A+I)}^n={(I+A)}^n$

$$\begin{gathered} =I+{}^{n}C_{1}A+{}^{n}C_2{A}^2+{}^{n}C_3{A}^3+...+{}^{n}C_n{A}^n \\ =I+({}^{n}C_1+{}^{n}C_2+{}^{n}C_3+...+{}^{n}C_n)A [from Eq.(i\left)\right] \\ =I+(2^n-1)A \end{gathered}$$

Hence, ${(A+I)}^n=I+(2^n-1)A, \forall n\in N$.