Question

Prove that 2n^3 + 3n^2 + n is divisible by 6 for every integer n > 1 by mathematical induction. (7 marks)

Answer 1

The answer is True

Explanation:

A mathematical induction consists in only 2 steps:

First step: Show the proposition is true for the first one valid integer number.

Second step: Show that if any one is true then the next one is true

Finally, if first step and second step are true, then the complete proposition is true.

So, given
`2*n^3+3*n^2+n`

First step: using and replacing n=2 (the first valid integer number >1)

`2*(2)^3 +3*(2)^2+2=30`

`(30)/(6) =5`

As the result is an integer number, so the first step is true.

Second step: using any next number,
`n+1`, let it replace

`2*(n+1)^3+3*(n+1)^2+(n+1)\\2*(n^3+3*n^2+3*n+1)+3*(n^2+2*n+1)^2+(n+1)\\2*n^3+6*n^2+6*n+2+3*n^2+6*n+3+n+1\\(2*n^3+3*n^2+n)+(6*n^2+12*n+6)`

As the First step is true, we know that

`2*n^3+3*n^2+n`
`=6*k`,

So let it replace in the previous expression

`6*k+6*(n^2+2*n+1)\\6*[k+(n^2+2*n+1)]`

Finally

`(6*[k+(n^2+2*n+1)])/(6) =k+(n^2+2*n+1)`

where the last expression is an integer number

So the second step is true, and the complete proposition is True