Question 21:

Prove the following by using the principle of mathematical induction for all n ∈N : x²ⁿ – y²ⁿ is divisible by x + y.

Answer

Let the given statement be P(n), i.e.,

P(n): x²ⁿ − y²ⁿ is divisible by x + y.

It can be observed that P(n) is true for n = 1.

This is so because x^{2 × 1} – y^{2 × 1} = x² – y² = (x + y) (x – y) is divisible by (x +y)

Let P(k) be true for some positive integer k, i.e.,

x^2k – y^2k is divisible by x + y.

∴x^2k – y^2k = m (x + y), where m ∈ N … (1)

We shall now prove that P(k + 1) is true whenever P(k) is true.

Consider

x^2(k+1) – y^{2(k + 1)}

=x^2k ∙ x² – y^2k ∙ y²

=x²(x^2k – y^2k + y^2k) – y^2k ∙y²

=x²{m(x + y) + y^2k} – y^2k ∙ y²              [Using (1)]

=m(x + y)x² + y^2k ∙ x² – y^2k ∙ y²

=m(x + y)x² + y^2k (x² – y²)

=m(x + y)x² + y^2k(x + y) (x – y)

=(x + y) {mx² + y^2k(x – y)}, which is factor of (x + y).

Thus, P(k + 1) is true whenever P(k) is true.

Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., n.

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