# If B, C are

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 = A2and 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.

Ak+1 = Bk(B + (k + 1)C)

To prove the given equation using mathematical induction, we have to show that Ak+2 = Bk+1(B + (k + 2)C).

We know Ak+2 = Ak+1 × A.

Ak+2 = [Bk(B + (k + 1)C)](B + C)

Ak+2 = [Bk+1 + (k + 1)BkC)](B + C)

Ak+2 = Bk+1(B + C) + (k + 1)BkC(B + C)

Ak+2 = Bk+1(B + C) + (k + 1)BkCB + (k + 1)BkC2

However, BC = CB and C2 = O.

Ak+2 = Bk+1(B + C) + (k + 1)BkBC + (k + 1)BkO

Ak+2 = Bk+1(B + C) + (k + 1)Bk+1C + O

Ak+2 = Bk+1(B + C) + Bk+1[(k + 1)C]

Ak+2 = Bk+1[(B + C) + (k + 1)C]

Ak+2 = Bk+1[B + (1 + k + 1)C]

Ak+2 = Bk+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.

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