Prove the statement by using the Principle of Mathematical Induction: For any natural number n, xn – yn is divisible by x – y, where x and y are any integers with x ≠ y.

Question

Prove the statement by using the Principle of Mathematical Induction:

For any natural number n, xⁿ – yⁿ is divisible by x – y, where x and y are any integers with x ≠ y.

Theorem

Solution

P(n) = xⁿ – yⁿ is divisible by x – y, x integers with x ≠ y.

So, substituting different values for n, we get,

P(0) = x⁰ – y⁰ = 0 Which is divisible by x − y.

P(1) = x − y Which is divisible by x − y.

P(2) = x² – y²

= (x + y)(x − y) Which is divisible by x − y.

P(3) = x³ – y³

= (x − y)(x² + xy + y²) Which is divisible by x − y.

Let P(k) = x^k – y^k be divisible by x – y;

So, we get,

⇒ x^k – y^k = a(x − y).

Now, we also get that,

⇒ P(k + 1) = x^k+1 – y^k+1

= x^k(x − y) + y(x^k − y^k)

= x^k(x − y) + y^a(x − y) Which is divisible by x − y.

⇒ P(k + 1) is true when P(k) is true.

Therefore, by Mathematical Induction,

P(n) xⁿ – yⁿ is divisible by x – y, where x integers with x ≠ y which is true for any natural number n.

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Principle of Mathematical Induction

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Chapter 4: Principle of Mathematical Induction - Exercise [Page 71]

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NCERT Exemplar Mathematics Exemplar [English] Class 11

Chapter 4 Principle of Mathematical Induction
Exercise | Q 8 | Page 71

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