<s>[INST] Arithmetic series relationship with difference of two consecutive cubes. Is this a thing? Excuse my dodgy notation and my write-up in general, this is the first proof I've done since leaving school a while back. Anywho, has anyone come across anything like this before? Read the whole thing, not just the arithmetic series stuff.
The sum of an ascending series, where a is the first integer, k is the common difference, and there are n terms.
$a+(a+k)+(a+2k)+⋯(a+(n-1)k)$
$$=na+[0+(n-1)k)+[1+(n-2)k]+[2+(n-3)k]+⋯$$
$$=na+\frac {n}{2}(n-1)k$$
$$=n\left[a+\frac{(n-1)k}{2}\right]$$
Difference between two consecutive cubes:
$2^3-1^3=7=(1×6)+1$
$3^3-2^3=19=(1+2)×6+1$
$4^3-3^3=37=(1+2+3)×6+1$
By observation, the difference between two consecutive cubes is equal to six times the sum of all integers between $0$ and the larger cubed integer, plus $1$.
Algebraically:
$$1+6\sum_{i=1}^x a_i=1+6{n\left[a+\frac{(n-1)k}{2}\right]}$$
Where $1$ is the first term and the common difference, and there are x terms.
$$=1+6{x\left[1+\frac{(x-1)}{2}\right]}$$ 
$$=1+6x+3x^2-3x$$
$$=3x^2+3x+1$$
$$=x^3+3x^2+3x+1-x^3$$
$$=(x+1)^3-x^3$$
Q.E.D
P.S. I always wanted to use Q.E.D but don't normally get the chance.
 [/INST] I think you can prove it directly :
$$(n+1)^3 - n^3 = 3 n^2 + 3 n + 1 = 3 n (n + 1) + 1 = 6 \frac{n (n + 1)}  2 + 1  $$
$\frac{n (n + 1)}  2$ being the sum of all integers from $1$ to $n$
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