If B, C are n rowed square matrices and if A = B + C,

Let $P(n)$ be the statement given by $P(n): A^{n+1}=B^{n}(B+(n+1) C)$.

For $n=1$, we have

$P(1): A^{2}=B(B+2 C)$

Here,

$\mathrm{LHS}=A^{2}$

$=(B+C)(B+C)$

$=B(B+C)+C(B+C)$

$=B^{2}+B C+C B+C^{2}$

$=B^{2}+2 B C$     $\left[\because B C=C B\right.$ and $\left.C^{2}=O\right]$

$=B(B+2 C)=\mathrm{RHS}$

Hence, the statement is true for $n=1$.

If the statement is true for

$n=k$, then $P(k): A^{k+1}=B^{k}(B+(k+1) C)$   $\ldots(1)$

For $P(k+1)$ to be true, we must have

$P(k+1): A^{k+2}=B^{k+1}(B+(k+2) C$

Now,

$A^{k+2}=A^{k+1} A$

$=\left[B^{k}(B+(k+1) C)\right](B+C) \quad$ [From eqn. (1) $]$

$=\left[B^{k+1}+(k+1) B^{k} C\right](B+C)$

$=B^{k+1}(B+C)+(k+1) B^{k} C(B+C)$

$=B^{k+2}+B^{k+1} C+(k+1) B^{k} C B+(k+1) B^{k} C^{2}$

$=B^{k+2}+B^{k+1} C+(k+1) B^{k} B C \quad\left[\because B C=C B\right.$ and $\left.C^{2}=0\right]$

$=B^{k+2}+B^{k+1} C+(k+1) B^{k+1} C$

$=B^{k+2}+(k+2) B^{k+1} C$

$=B^{k+1}[B+(k+2) C]$

So the statement is true for $n=k+1$.

Hence, by the principle of mathematical induction, $P(n)$ is true for all $n \in N$.