Find $A^n$, if $n$ is Natural number Find eqation for:$\ A^n$ when:
$$A =\left( \begin{array}{cc}
a & 1 & 0\\
0 & a & 1\\
0 & 0 & a\end{array} \right)$$
I calculated $\ A^2$ $\ A^3$ and compared entries:
At the end I've got something like:
$$A^n =\left( \begin{array}{cc}
a^n & na^{n-1} & ???\\
0 & a^n & na^{n-1}\\
0 & 0 & a^n\end{array} \right)$$
So I'm struggling with estimating value of the last entry.
From my calculations I've got respectively: $\ 0 $ $\ 1 $ $\ 3a $. How I can find the equation for this sequence?

$$A^2 = \left(
\begin{array}{ccc}
 a^2 & 2 a & 1 \\
 0 & a^2 & 2 a \\
 0 & 0 & a^2 \\
\end{array}
\right)$$
$$A^3 = \left(
\begin{array}{ccc}
 a^3 & 3 a^2 & 3 a \\
 0 & a^3 & 3 a^2 \\
 0 & 0 & a^3 \\
\end{array}
\right)$$
$$A^4 = \left(
\begin{array}{ccc}
 a^4 & 4 a^3 & 6 a^2 \\
 0 & a^4 & 4 a^3 \\
 0 & 0 & a^4 \\
\end{array}
\right)$$
$$A^5 = \left(
\begin{array}{ccc}
 a^5 & 5 a^4 & 10 a^3 \\
 0 & a^5 & 5 a^4 \\
 0 & 0 & a^5 \\
\end{array}
\right)$$
Now the sequence in the top right element:
$$1, 3, 6, 10, ...$$
Doesn't remember you about the binomial expansion? :)
