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Prove by method of induction \(\begin{pmatrix}1&2\\0&1\end{pmatrix}^n=\begin{pmatrix}1&2n\\0&1\end{pmatrix}\forall\in N\) 

[(1,2),(0,1)]n = [(1, 2n), (0, 1)] ∀ ∈ N

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Let P(n) = \(\begin{bmatrix}1&2\\0&1\end{bmatrix}^n=\begin{bmatrix}1&2n\\0&1\end{bmatrix}\), for all n ∈ N

Step IV:

From all the steps above, by the principle of mathematical induction, P(n) is true for all n ∈ N.

\(\therefore\) \(\begin{bmatrix}1&2\\0&1\end{bmatrix}^n=\begin{bmatrix}1&2n\\0&1\end{bmatrix}\), for all n ∈ N.

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