Question

Use the Principle of Mathematical Induction to show that n^3+2n+6 is divisible by 3 for all natural numbers n.

Answer 1

The proposition: n^3+2n+6 is divisible by 3 for all natural numbers n

It's a three step process:
1. basis: for n=1, n^3+2n+6=1+2+6=9 =0 mod(3)
2. Assume the proposition is true for k, i.e. 3 | k^3+2k+6
3. Induction step:
For n=k+1, we consider
(k+1)^3+2(k+1)+6
=(k^3+3k^2+3k+1)+2k+2+6
=(k^3+3k+6) + 3k^2+1+2
=(k^3+3k+6) + 3(k^2+1)
Now examine the expression.
The first term has already been assumed to be true, so it's divisible by 3.
The second term clearly has a factor of three, so it is also divisible by 3.
We know that the sum of two quantities is divisible by 3 when each is divisible by three. Therefore the assumption is also true for n=k+1.

Thus the proposition is true for all positive values of k.