Question

Using the Mathematical induction, show that for any natural number n, x²ⁿ − y²ⁿ is divisible by x + y

Answer 1

Let P(n) = x²ⁿ – y²ⁿ is divisible by (x + y)

For n = 1

P(1) = x^{2 × 1} – y^{2 × 1} is divisible by (x + y)

⇒ (x + y) (x – y) is divisible by (x + y)

∴ P(1) is true

Let P(n) be true for n = k

∴ P(k) = x^2k – y^2k is divisible by (x + y)

⇒ x^2k – y^2k = λ(x + y)  ......(i)

For n = k + 1

⇒ P(k + 1) = `x^(2("k" + 1)) – y^(2("k" + 1))` is divisible by (x + y)

Now `x^(2("k" + 2)) – y^(2("k" + 2))` 

= `x^(2"k" + 2) – x^(2k)y^2 + x^(2k)y^2 – y^(2k) + 2`

= `x^(2k)*x^2 – x^(2k)y^2 + x^(2k)y^2 – y^(2k)y^2`

= `x^(2k) (x^2 – y^2) + y^2λ (x + y)`  ......[Using (i)]

⇒ `x^(2"k" + 2) – y^(2"k" + 2)` is divisible by (x + y)

∴ P(k + 1) is true.

Thus P(k) is true

⇒ P(k + 1) is true.

Hence by principle of mathematical induction,

P(n) is true for all n ∈ N