By the principle of mathematical induction, provean – bn is divisibl

1.	By the principle of mathematical induction, provean – bn is divisible by a – b, for all n ∈ N.
Answer» 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.

By the principle of mathematical induction, provean – bn is divisible by a – b, for all n ∈ N.

Answer»

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.