<s>[INST] Finding $A^n$ for a matrix I have a matrix $$
A =
\left[ {\begin{array}{cc}
 1 & c  \\
 0 & d  \\
 \end{array} } \right]
$$
with $c$ and $d$ constant. I need to find $A^n$ ($n$ positive) and then need to prove that formula using induction.
I would like to check that the formula I derived is correct: 
$$
A^n =
\left[ {\begin{array}{cc}
 1 & c^{n-2}(dc + c)  \\
 0 & d^n  \\
 \end{array} } \right]
$$
If this is correct, how can I prove this? I suppose I can write $A^{n+1} = A^n A$, which would be
$$
\left[ {\begin{array}{cc}
 1 & c^{n-2}(dc + c)  \\
 0 & d^n  \\
 \end{array} } \right]
\left[ {\begin{array}{cc}
 1 & c  \\
 0 & d  \\
 \end{array} } \right]

$$
But then what would I do?
Thanks.
 [/INST] Mathematical induction is the way to go, but first you want to have a "target." I'm not sure what you did, but I think you confused $c$ and $d$ somewhere along the line in your calculations. Let's see a first few values:
$$\begin{align*}
A&= \left(\begin{array}{cc}1&c\\0&d\end{array}\right)\\
A^2 &= \left(\begin{array}{cc}1&c\\0&d\end{array}\right)\left(\begin{array}{cc}1&c\\0&d\end{array}\right) = \left(\begin{array}{cc}1 & c+cd\\0&d^2\end{array}\right).\\
A^3 &= \left(\begin{array}{cc}
1&c+cd\\
0 &d^2\end{array}\right) \left(\begin{array}{cc}
1 & c\\
0 & d\end{array}\right) = \left(\begin{array}{cc}
1 & c+cd+cd^2\\
0 & d^3
\end{array}\right).\\
A^4 &= \left(\begin{array}{cc}
1&c+cd+cd^2\\
0 & d^3
\end{array}\right)\left(\begin{array}{cc}
1 & c\\0 & d\end{array}\right) = \left(\begin{array}{cc}
1 & c + cd + cd^2 + cd^3\\
0 & d^4
\end{array}\right).
\end{align*}$$
Okay, that suggests the pattern: 
Conjecture. For every positive integer $n$, 
$$A^n = \left(\begin{array}{cc}
1 & c(1+d+d^2+\cdots + d^{n-1})\\
0 & d^n
\end{array}\right).$$
To prove the conjecture, we use mathematical induction. Prove the formula is true for $n=1$, and then, assuming the formula holds for $k$, prove it holds for $k+1$. We have:
Basis. For $n=1$, we have
$$A = \left(\begin{array}{cc}1 & c\\0 & d\end{array}\right),$$
so the formula holds.
Inductive step. Show that if the formula holds for $k$, then it also holds for $k+1$.
Induction hypothesis. Assume the result holds for $k$; that is, that
$$A^k = \left(\begin{array}{cc}
1 & c(1+d+\cdots+d^{k-1})\\0 & d^k
\end{array}\right).$$
Now we have:
$$\begin{align*}
A^{k+1} & = A^kA\\
&= \left(\begin{array}{cc}
1 & c(1+d+\cdots + d^{k-1})\\0 & d^k\end{array}\right)
\left(\begin{array}{cc} 1 & c\\0 & d\end{array}\right)\\
&= \left(\begin{array}{cc}
1 & c + cd(1+d+\cdots + d^{k-1})\\
0 & d^{k+1}\end{array}\right)\\
&= \left(\begin{array}{cc}
1 & c(1 + d(1+d+\cdots + d^{k-1}))\\
0 & d^{k+1}\end{array}\right)\\
&= \left(\begin{array}{cc}
1 & c(1+d+d^2+\cdots + d^k)\\
0 & d^{k+1}\end{array}\right).
\end{align*}$$
That is exactly the formula we have evaluated at $k+1$. Therefore, if the formula holds for $k$, then it holds for $k+1$ as well.
So: the formula holds for $n=1$; and if it holds for $n=k$, then it also holds for $n=k+1$. By Mathematical Induction, we conclude that the formula holds for all $n$.
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