a^n – b^n is divisible by a – b, for all n ∈ N.

Question

By the principle of mathematical induction, prove

aⁿ – bⁿ is divisible by a – b, for all n ∈ N.

Answer 1

Let P(n) denote the statement aⁿ – bⁿ is divisible by a – b. 

Put n = 1. Then P(1) is the statement: a¹ – b¹ = a – b is divisible by a – b 

∴ P(1) is true. Now assume that the statement be true for n = k 

(i.e.,) assume P(k) be true, (i.e.,) a^k – b^k is divisible by (a – b) be true. 

⇒ \(\frac{a^k-b^k}{a-b}\) = m (say) where m ∈ N

⇒ a^k – b^k = m(a – b) 

⇒ a^k = b^k + m(a – b) … (1) 

Now to prove P(k + 1) is true, (i.e.,) to prove: a^{k + 1} – b^{k + 1} is divisible by a – b 

Consider a^{k + 1} – b^{k + 1} = a^k.a – b^k.b 

= [b^k + m(a – b)] a – b^k.b [∵ a^k = bm + k(a – b)] 

= b^k.a + am(a – b) – b^k.b 

= b^k.a – b^k.b + am(a – b) 

= b^k(a – b) + am(a – b) 

= (a – b) (b^k + am) is divisible by (a – b) 

∴ P(k + 1) is true. 

By the principle of Mathematical induction. P(n) is true for all n ∈ N.

∴ aⁿ – bⁿ is divisible by a – b for n ∈ N.