Step 1: Understanding the Concept:
This question requires verifying several statements involving matrix multiplication and exponentiation. We will evaluate each statement individually.
Step 2: Detailed Explanation:
Statement (A):
\[
B^n = \begin{pmatrix} 1 & nq \\ 0 & 1 \end{pmatrix}
\]
For \(n=2\):
\[
B^2 = \begin{pmatrix} 1 & q \\ 0 & 1 \end{pmatrix}
\begin{pmatrix} 1 & q \\ 0 & 1 \end{pmatrix}
= \begin{pmatrix} 1 & 2q \\ 0 & 1 \end{pmatrix}.
\]
This matches the formula. By induction, it holds true.
Statement (A) is correct.
Statement (B):
\[
A^n = \begin{pmatrix} p^n & q\frac{p^n-1}{p-1} \\ 0 & 1 \end{pmatrix}, \; (p \neq 1)
\]
For \(n=2\):
\[
A^2 = \begin{pmatrix} p & q \\ 0 & 1 \end{pmatrix}
\begin{pmatrix} p & q \\ 0 & 1 \end{pmatrix}
= \begin{pmatrix} p^2 & q(p+1) \\ 0 & 1 \end{pmatrix}.
\]
Formula gives:
\[
q\frac{p^2-1}{p-1} = q(p+1).
\]
Matches perfectly. This comes from the geometric progression in the top-right entry.
Statement (B) is correct.
Statement (C):
\[
AB = \begin{pmatrix} p & pq+q \\ 0 & 1 \end{pmatrix}
\]
\[
AB = \begin{pmatrix} p & q \\ 0 & 1 \end{pmatrix}
\begin{pmatrix} 1 & q \\ 0 & 1 \end{pmatrix}
= \begin{pmatrix} p & pq+q \\ 0 & 1 \end{pmatrix}.
\]
Statement (C) is correct.
Statement (D):
\[
B^{n-1} = \begin{pmatrix} 1 & (n+1)q \\ 0 & 1 \end{pmatrix}
\]
But from (A):
\[
B^k = \begin{pmatrix} 1 & kq \\ 0 & 1 \end{pmatrix}.
\]
So,
\[
B^{n-1} = \begin{pmatrix} 1 & (n-1)q \\ 0 & 1 \end{pmatrix}.
\]
The given version is wrong.
Statement (D) is incorrect.
Statement (E):
\[
AB^n = \begin{pmatrix} p & (np+1)q \\ 0 & 1 \end{pmatrix}
\]
Using (A):
\[
AB^n = \begin{pmatrix} p & q \\ 0 & 1 \end{pmatrix}
\begin{pmatrix} 1 & nq \\ 0 & 1 \end{pmatrix}
= \begin{pmatrix} p & npq+q \\ 0 & 1 \end{pmatrix}
= \begin{pmatrix} p & (np+1)q \\ 0 & 1 \end{pmatrix}.
\]
Statement (E) is correct.
Step 3: Final Answer:
The correct statements are: (A), (B), (C), and (E).
Only (D) is incorrect.
Correct Choice: Option 1.