Given,

A = B + C, BC = CB and C² = O. 

We need to prove that,

Aⁿ⁺¹ = Bⁿ(B + (n + 1)C). 

We will prove this result using the principle of mathematical induction.

Step 1: 

When n = 1, we have Aⁿ⁺¹ = A¹⁺¹ 

⇒ Aⁿ⁺¹ = B¹(B + (1 + 1)C) 

∴ Aⁿ⁺¹ = B(B + 2C)

For the given equation to be true for n = 1,

Aⁿ⁺¹ must be equal to A². 

It is given that,

A = B + C 

And we know, 

A² = A × A.

⇒ A² = (B + C)(B + C) 

⇒ A² = B(B + C) + C(B + C) 

⇒ A² = B2 + BC + CB + C² 

However, 

BC = CB and C² = O. 

⇒ A² = B² + CB + CB + O 

⇒ A² = B² + 2CB 

∴ A² = B(B + 2C) 

Hence, 

Aⁿ⁺¹ = A² and the equation is true for n = 1.

Step 2 : 

Let us assume the equation true for some n = k, where k is a positive integer. 

⇒ A^k+1 = B^k (B + (k + 1)C) 

To prove the given equation using mathematical induction, we have to show that,

A^k+2 = B^k+1(B + (k + 2)C). 

We know,

A^k+2 = A^k+1 × A. 

⇒ A^k+2 = [B^k(B + (k + 1)C)](B + C) 

⇒ A^k+2 = [B^k+1 + (k + 1)B^kC)](B + C) 

⇒ A^k+2 = B^k+1(B + C) + (k + 1)B^kC(B + C) 

⇒ A^k+2 = B^k+1(B + C) + (k + 1)B^kCB + (k + 1)B^kC² 

However, 

BC = CB and C² = O. 

⇒ A^k+2 = B^k+1(B + C) + (k + 1)B^kBC + (k + 1)B^kO 

⇒ A^k+2 = B^k+1(B + C) + (k + 1)B^k+1C + O 

⇒ A^k+2 = B^k+1(B + C) + B^k+1[(k + 1)C] 

⇒ A^k+2 = B^k+1[(B + C) + (k + 1)C] 

⇒ A^k+2 = B^k+1[B + (1 + k + 1)C] 

∴ A^k+2 = B^k+1[B + (k + 2)C] 

Hence, 

The equation is true for n = k + 1 under the assumption that it is true for n = k. 

Therefore, 

By the principle of mathematical induction, the equation is true for all positive integer values of n. 

Thus, 

Aⁿ⁺¹ = Bⁿ(B + (n + 1)C) for every n ϵ N.