Prove algebraically that n^3 +3n -1 is odd for all positive integers n

There are 2 possible cases.

First when n is even, then let n = 2k where k is a positive integer.
Substituting n = 2k gives us:
(2k)³ + 3(2k) -1
= 8k³ + 6k - 1
[We are trying to prove that it is odd so try and write it in the form 2m + 1 or 2m - 1]
= 2(4k³ + 3k) - 1
Since 2(4k³ + 3k) is a multiple of 2 it is even so by taking away 1 makes it odd, therefore when n=2k the expression is odd.

The second case is when n is odd, then let n=2k + 1 where k is a positive integer (including 0).
Substituting n=2k + 1 gives us:
(2k+1)³ + 3(2k+1) -1
[use the binomial theorem to expand (2k+1)³]
(2k)³ + 3(2k)²(1) + 3(2k)(1)² + (1)³ + 6k + 3 - 1
=8k³ + 12k² + 6k + 1 + 6k + 2
=8k³ + 12k² + 12k + 3
=2(4k³ + 6k² + 6k) + 3
This is an even number + 3 which gives us and odd number, therefore when n=2k+1 the expression is odd.
Therefore we can conclude that for all n, n³ + 3n - 1 is odd.