By the principle of mathematical induction, prove 32n – 1 is divis

1.	By the principle of mathematical induction, prove 32n – 1 is divisible by 8, for all n ∈ N.
Answer» Let P(n) denote the statement 3²ⁿ – 1 is divisible by 8 for all n ∈ N Put n = 1 P(1) is the statement 3²⁽¹⁾ – 1 = 3² – 1 = 9 – 1 = 8, which is divisible by 8 ∴ P(1) is true. Assume that P(k) is true for n = k. i.e., 3^2k – 1 is divisible by 8 to be true. Let 3^2k – 1 = 8m To prove P(k + 1) is true. i.e., to prove 3^{2(k + 1)} – 1 is divisible by 8 Consider 3^{2(k + 1)} – 1 = 3^{2k + 2} – 1 = 3^2k.3² – 1 = 3^2k(9) – 1 = 3^2k(8 + 1) – 1 = 3 x 8 + 3 x 1 – 1 = 3^2k(8) + 3^2k – 1 = 3^2k(8) + 8m (∵ 3^2k – 1 = 8m) = 8(3^2k + m), which is divisible by 8. ∴ P(k + 1) is true wherever P(k) is true. ∴ By principle of Mathematical Induction, P(n) is true for all n ∈ N.

By the principle of mathematical induction, prove 32n – 1 is divisible by 8, for all n ∈ N.

Answer»

Let P(n) denote the statement 3²ⁿ – 1 is divisible by 8 for all n ∈ N

Put n = 1

P(1) is the statement 3²⁽¹⁾ – 1 = 3² – 1 = 9 – 1 = 8, which is divisible by 8

∴ P(1) is true.

Assume that P(k) is true for n = k.

i.e., 3^2k – 1 is divisible by 8 to be true.

Let 3^2k – 1 = 8m

To prove P(k + 1) is true.

i.e., to prove 3^{2(k + 1)} – 1 is divisible by 8

Consider 3^{2(k + 1)} – 1 = 3^{2k + 2} – 1

= 3^2k.3² – 1

= 3^2k(9) – 1

= 3^2k(8 + 1) – 1

= 3 x 8 + 3 x 1 – 1

= 3^2k(8) + 3^2k – 1

= 3^2k(8) + 8m (∵ 3^2k – 1 = 8m)

= 8(3^2k + m), which is divisible by 8.

∴ P(k + 1) is true wherever P(k) is true.

∴ By principle of Mathematical Induction, P(n) is true for all n ∈ N.