mathmatical induction (F.5)

1)When k = 1 , 5 - 3 - 2 = 0 is divisible by 15 .Assuming that 5²ⁿ⁻¹ - 3²ⁿ⁻¹ - 2²ⁿ⁻¹ is divisible by 15 ,
i.e. 5²ⁿ⁻¹ - 3²ⁿ⁻¹ - 2²ⁿ⁻¹ = 15m , where m is a integer.When k = n+1 , 5²⁽ⁿ⁺¹⁾⁻¹ - 3²⁽ⁿ⁺¹⁾⁻¹ - 2²⁽ⁿ⁺¹⁾⁻¹ = 5²ⁿ⁺¹ - 3²ⁿ⁺¹ - 2²ⁿ⁺¹ = (25) 5²ⁿ⁻¹ - (9) 3²ⁿ⁻¹ - (4) 2²ⁿ⁻¹= 4(5²ⁿ⁻¹ - 3²ⁿ⁻¹ - 2²ⁿ⁻¹) + (21) 5²ⁿ⁻¹ - (5)3²ⁿ⁻¹ = 4(15m) + (105)5²ⁿ⁻² - (15)3²ⁿ⁻² = 15(4m + (7)5²ⁿ⁻² - 3²ⁿ⁻²) is divisible by 15.∴ By the principle of mathematical induction , 5²ᵏ⁻¹ - 3²ᵏ⁻¹ - 2²ᵏ⁻¹ is divisible by 15 , for all positive integers k.

2)When n = 1 , (3+1)7 - 1 = 27 is divisible by 9.Assuming that (3k+1)7ᵏ - 1 is divisible by 9 ,
i.e. (3k+1)7ᵏ - 1 = 9m , where m is a positive integer.When n = k+1 , (3(k+1) + 1)7ᵏ⁺¹ - 1= (3(k+1) + 1)7ᵏ⁺¹ - 1 - ( (3k+1)7ᵏ - 1 ) + ( (3k+1)7ᵏ - 1 ) = 7(3k+4)7ᵏ - (3k+1)7ᵏ + 9m= (18k+27) 7ᵏ + 9m= 9 ( (2k+3) 7ᵏ + m ) is divisible by 9.∴ By the principle of mathematical induction , (3n+1)7ⁿ - 1 is divisible by 9 ,
for all positive integers n.