Q.Prove by method of induction $\begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}^n = \begin{pmatrix} 1 & 2n \\ 0 & 1 \end{pmatrix}$, $\forall n \in N$.

Let $A=\begin{pmatrix}1&2\\0&1\end{pmatrix}$ and $P(n):A^n=\begin{pmatrix}1&2n\\0&1\end{pmatrix}$. Base: $n=1$: $A^1=A=\begin{pmatrix}1&2\\0&1\end{pmatrix}$, matching the formula at $n=1$; holds. Hypothesis: assume $A^k=\begin{pmatrix}1&2k\\0&1\end{pmatrix}$. Step: $A^{k+1}=A^k\cdot A=\begin{pmatrix}1&2k\0&1\end{pmatrix}\begin{pmatrix}1&2\0&1\end{pmatrix}=\begin{pmatrix}1\cdot1+2k\cdot0 & 1\cdot2+2k\cdot1\0\cdot1+1\cdot0 & 0\cdot2+1\cdot1\end{pmatrix}=\begin{pmatrix}1&2+2k\0&1\end{pmatrix} …

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