Induction Proof for A^n = 1 2^nProve your formula by mathematical induction.

[Mark44]

@Robb, as already mentioned, for an induction proof you need to

1. Establish a base case (e.g., with n = 1).
2. Assume that the proposition is true if n = k.
3. Show that if the proposition is true for n = k, it must also be true for n = k + 1.

The base case is trivial in this problem.
For step 2, it's reasonable to assume that for n = k, $A^k = \begin{bmatrix} 1 & 2k \\ 0 & 1\end{bmatrix}$
For step 3, show, using the assumption in step 2, that $A^{k + 1} = \begin{bmatrix} 1 & 2(k + 1) \\ 0 & 1\end{bmatrix}$
Your work should start with $A^{k + 1} = \dots$.
That's it!

 

[Robb]

@Robb, as already mentioned, for an induction proof you need to

1. Establish a base case (e.g., with n = 1).
2. Assume that the proposition is true if n = k.
3. Show that if the proposition is true for n = k, it must also be true for n = k + 1.

The base case is trivial in this problem.
For step 2, it's reasonable to assume that for n = k, $A^k = \begin{bmatrix} 1 & 2k \\ 0 & 1\end{bmatrix}$
For step 3, show, using the assumption in step 2, that $A^{k + 1} = \begin{bmatrix} 1 & 2(k + 1) \\ 0 & 1\end{bmatrix}$
Your work should start with $A^{k + 1} = \dots$.
That's it!

Now that makes sense! As always, the help is much appreciated!