The sum of all natural numbers smaller than and including x is equal to

(x+1)(x/2)

The sum of all even numbers up to and including x is that minus (n/2)² so

[(x+1)(x/2)]-(x/2)²

That would mean the sum of all odd numbers under x is equal to

(x/2)²

or sum of all odd including x (if x is odd of course) is

[(x+1)/2]²

Since the sum of all even numbers up to x is the sum of all numbers minus the sum of all odd numbers.

[-(x/2)² +2x +1]/2 (another way of writing the sum of all evens under and including a number) looks suspiciously polynomial. I want to go further.

Edit:

Interestingly, the sum of all evens under and including x is larger than the sum of all odds under x by half of x. So

[(x+1)(x/2)-(x/2)²]-(x/2) = (x/2)²

So another formula for sum of all evens is

(x/2)²+(x/2)

So,

2[(x/2)²]+(x/2) = (x+1)(x/2)

The left is the sum of all odds plus the sum of all evens under and including x, the right is the original formula we started with, sum of all natural numbers under and including x. Since they both give us the total sum for all natural numbers up to and including x, the left hand side is a different formula giving us the same result!