By the principle of mathematical induction, prove52n – 1 is divisibl

1.	By the principle of mathematical induction, prove52n – 1 is divisible by 24, for all n ∈ N.
Answer» Let P(n) be the proposition that 5²ⁿ – 1 is divisible by 24. For n = 1, P(1) is: 5² – 1 = 25 – 1 = 24, 24 is divisible by 24. Assume that P(k) is true. i.e., 5^2k – 1 is divisible by 24 Let 5^2k – 1 = 24m To prove P(k + 1) is true. i.e., to prove 5^{2(k + 1)} – 1 is divisible by 24. P(k): 5^2k – 1 is divisible by 24. P(k + 1) = 5^{2(k + 1)} – 1 = 5^2k.5² – 1 = 5^2k(25) – 1 = 5^2k(24 + 1) – 1 = 24.5^2k + 5^2k – 1 = 24.5^2k + 24m = 24 [5^2k + 24] Which is divisible by 24 ⇒ P(k + 1) is also true. Hence by mathematical induction, P(n) is true for all values n ∈ N.

By the principle of mathematical induction, prove52n – 1 is divisible by 24, for all n ∈ N.

Answer»

Let P(n) be the proposition that 5²ⁿ – 1 is divisible by 24.

For n = 1, P(1) is: 5² – 1 = 25 – 1 = 24, 24 is divisible by 24.

Assume that P(k) is true.

i.e., 5^2k – 1 is divisible by 24

Let 5^2k – 1 = 24m

To prove P(k + 1) is true.

i.e., to prove 5^{2(k + 1)} – 1 is divisible by 24.

P(k): 5^2k – 1 is divisible by 24.

P(k + 1) = 5^{2(k + 1)} – 1

= 5^2k.5² – 1

= 5^2k(25) – 1

= 5^2k(24 + 1) – 1

= 24.5^2k + 5^2k – 1

= 24.5^2k + 24m

= 24 [5^2k + 24]

Which is divisible by 24 ⇒ P(k + 1) is also true.

Hence by mathematical induction, P(n) is true for all values n ∈ N.