Prove that if n is a positive integer then n3 2n is a multi

Prove that if n is a positive integer, then n³ + 2n is a multiple of 3. ( use induction) (Discrete Mathematics)

(I got this wrong on a test and would really like to understand how to do it)

Solution

To prove : n^3 + 2n is a multiple of 3.

Step 1 : If n = 0 , then n^{^3} + 2n = 0^3 + 2×0 = 0. So it\'s a multiple of 3.

Step 2 Induction: Assume that for some positive integre number n, n^{^3} + 2n is a multiple of 3.

step 3 Induction Hypothesis: To prove this for n+1, we\'ll try to express ( n + 1 )^3 + 2( n + 1 ) in terms of n^3 + 2n and use the induction hypothesis.

( n + 1 )^3 + 2( n + 1 ) = ( n^3 + 3n^2 + 3n + 1 ) + ( 2n + 2 )

= ( n^3 + 2n ) + ( 3n^3 + 3n + 3 )

= ( n^3 + 2n ) + 3( n^2 + n + 1 )

the second term is a multiple of 3 we just need to think about the first term ( n^3 + 2n )

= n^3 - n +3n = n(n^2 - 1) + 3n

= n(n+1)(n-1) + 3n

= (n-1)n(n+1) + 3n

now (n-1) , n , (n+1) are 3 consecutive integers and we know that the product of any three consecutive integers is a multiple of 3 and to this number we are adding 3n which is a mutiple of 3 as well

=> their sum will be a multiple of 3 as well

hence n^3 + 2n is a multiple of 3

which is divisible by 3, because ( n³ + 2n ) is divisible by 3 by the induction hypothesis.