mathematica induction

Question

 

8ⁿ-3ⁿ  is divisble by 5 

Answer 1

1. The statement is true for n=1, since 8¹-3¹=5 is divisible by 5.

2. Assume the statement is true for a fixed but arbitrary n=k:

8^k-3^k is divisible by 5. (induction hypothesis)

 

Then

8^k+1-3^k+1 = 8*8^k-3*3^k = (5+3)*8^k-3*3^k = 5*8^k+3*(8^k-3^k)

 

The first term is divisible by 5 because it contains a factor of 5. The second term is divisible by 5 because of our induction hypotheses. Therefore, 8^k+1-3^k+1 is divisible by 5. (induction step)

 

Then, by the principle of mathematical induction, 8ⁿ-5ⁿ is divisible by 5 for all integers n=1,2,3,...

Answer 2

8¹-3¹=8-3=5 is divisible by 5 - so it is valid for n=1

8ⁿ⁺¹-3ⁿ⁺¹=8×8ⁿ-8×3ⁿ+8×3ⁿ-3×3ⁿ=8(8ⁿ-3ⁿ)+3ⁿ(8-3)=8(8ⁿ-3ⁿ)+3ⁿ×5,  the first term divisible by 5 by hypothesis, the second term clearly so.  This proof exposes the second big secret of mathematics, the first being multiply by one, the second is add zero, underlined.