Question
If \(A = B + C\), where \(BC = CB\) and \(C^2 = O\), prove that for every \(n \in \mathbb{N}\), \[ A^{n+1} = B^n (B + (n+1)C). \]
Solution (Mathematical Induction)
Step 1: Base Case (n = 1)
\[ A^2 = (B + C)^2 = B^2 + BC + CB + C^2 \] Since \(BC = CB\) and \(C^2 = 0\), \[ A^2 = B^2 + 2BC = B(B + 2C) \] ✔ True for \(n=1\)Step 2: Assume for \(n = k\)
\[ A^{k+1} = B^k (B + (k+1)C) \]Step 3: Prove for \(n = k+1\)
\[ A^{k+2} = A^{k+1} \cdot A \] \[ = B^k (B + (k+1)C)(B + C) \]Step 4: Expand
\[ = B^k [B^2 + BC + (k+1)CB + (k+1)C^2] \] Using \(BC = CB\), \(C^2 = 0\): \[ = B^k [B^2 + (k+2)BC] \]Step 5: Factor
\[ = B^k B(B + (k+2)C) = B^{k+1}(B + (k+2)C) \]Step 6: Conclusion
✔ True for \(k+1\) \[ \Rightarrow A^{n+1} = B^n(B + (n+1)C) \]Final Result
\[
A^{n+1} = B^n (B + (n+1)C)
\]
Hence proved.