maths-MI

When n = 1 , (5¹)(4(1) - 1) + 1 = 16 is divisible by 16 .
Assuming that 5ᵏ (4k - 1) + 1 is divisible by 16 ,
i.e. 5ᵏ (4k - 1) + 1 = 16m , where m is a positive integer.
When n = k+1 , 5ᵏ⁺¹ (4(k+1) - 1) + 1= 5ᵏ⁺¹ (4(k+1) - 1) + 1 - ( 5ᵏ (4k - 1) + 1 ) + ( 5ᵏ (4k - 1) + 1 ) = 5ᵏ⁺¹ (4k + 3) - 5ᵏ (4k - 1) + ( 5ᵏ (4k - 1) + 1 )= 5ᵏ (20k + 15 - 4k + 1) + ( 5ᵏ (4k - 1) + 1 )= 5ᵏ (16k + 16) + ( 5ᵏ (4k - 1) + 1 )= 5ᵏ 16(k + 1) + 16m= 16 (5ᵏ (k + 1) + m) is divisible by 16.
∴ By the principle of mathematical induction , 5ⁿ (4n - 1) is divisible by 16 ,
for all positive integers n.

2012-04-24 13:46:53 補充：
∴ By the principle of mathematical induction , 5ⁿ (4n - 1) + 1 is divisible by 16 ,
for all positive integers n.