Summing a polynomial '0' thru 'n'

This writeup occured to me when I found a way to sum any polynomial from '0' thru 'n'. This all started when I needed to find a way to sum all the squares from '0' thru 'n'. I knew that the sum of numbers (0,1,2...) is given by:

(n(n + 1))/2

I just needed an equation to sum 0,1,4,9,16.... Now with this task in mind I started playing with squares and found some really cool patterns among them, but alas, not an equation.

But then something struck me. Now if I were to assume that the sum was given by a third-degree polynomial (which was an assumption based on the fact that the sum of numbers is given by a second-degree equation and the power of all numbers is 1, so the equation should be one greater degree than the power of the numbers) the difference of successive terms should be the square of that number. Mathematically:

(a(n³) + b(n²) + c(n) + d) – (a(( n - 1)³) + b(( n - 1)²) + c(( n - 1)) + d) ≡ n²; where the equation for the sum was ax³ + bx² + cx + d

Now obviously this has to be an identity because it would hold true for ALL values (the fact that it is an identity is important).

Now let us use that identity to get our equation!

(a(n³ + 3n² + 3n + 1) + b(n² + 2n + 1) + c(n + 1) + d) - (an³ + bn² + cn +d) ≡ n²

⇒ 3an² + (3a + 2b)n + a + b + c ≡ (n + 1)²

⇒ 3an² + (3a + 2b)n + a + b + c ≡ n² + 2n + 1

Now since it's an identity, we can equate powers.

3a + 2b = 2

a + b + c = 1

which gives:

a = 1/3; b = 1/2 ; c = 1/6

Now our equation shall be:

1/3(n³) + 1/2(n²) + 1/6 (n)

And now let's put some values...say we want to add 1,4,9,16,25,36. So n = 6, and from the calculator the value is 91.

Using our equation the value = 1/3(6³) + 1/2(6²) + 1/6(6)

⇒ 72 + 18 + 1

⇒ 91!!!!!!

So, our equation works! Which also proves that my initial assumption was correct that the equation we needed was a third-degree polynomial.

Now, having done it, I came up with a conjecture, which I personally don't need the feel to prove, because I know in my heart that it is true.(I am pretty sure it is also present in many many math books...)

So the conjecture is that for any polynomial (in our case it was y = x²), the sum from '0' thru 'n' is given by a polynomial of one greater degree. To find that polynomial, all you need to do is take its values at 'n' and 'n-1' (or 'n+1' and 'n') and equate its difference to y(n) (or y(n+1) if you used the 'n+1' and 'n' difference) where y(x) is the polynomial whose sum we need to find. Now you shall get simultaneous equations in form of a triangle (they are mighty easier to solve) and then you get your polynomial for the sum!