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If $A =$ $\left( {\begin{array}{*{20}{c}}1&a\\0&1\end{array}} \right)$ , then $A^n$ (where $n \in N$) equals
$\left( {\begin{array}{*{20}{c}}1&{na}\\0&1\end{array}} \right)$
$\left( {\begin{array}{*{20}{c}}1&{{n^2}a}\\0&1\end{array}} \right)$
$\left( {\begin{array}{*{20}{c}}1&{na}\\0&0\end{array}} \right)$
$\left( {\begin{array}{*{20}{c}}n&{na}\\0&n\end{array}} \right)$
Solution
We have $A^2 =$ $\left( {\begin{array}{*{20}{c}}1&a\\0&1\end{array}} \right)$ $\left( {\begin{array}{*{20}{c}}1&a\\0&1\end{array}} \right)$ $=$ $\left({\begin{array}{*{20}{c}}1&{2a}\\0&1\end{array}} \right)$
$A^3$ $= A^2A$ $=$ $\left( {\begin{array}{*{20}{c}}1&{2a}\\0&1\end{array}}\right)$ $\left( {\begin{array}{*{20}{c}}1&a\\0&1\end{array}} \right)$ $=$ $\left({\begin{array}{*{20}{c}}1&{3a}\\0&1\end{array}} \right)$
In general by induction, $A^n =$ $\left( {\begin{array}{*{20}{c}}1&{na}\\0&1\end{array}} \right)$ , $\forall \, n \in N$