Answer :

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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