This is a well known series, given by $y_k= 1/6\, k(k+1)(2k+1)$. There are several methods to prove it. One of the standard approaches is by induction. A simple and intuitive method is the following:
1 - write the sequence of differences between consecutive terms: $1, 4, 9, 16...$;
2 - write the sequence of differences between consecutive terms of the previous sequence: $1, 3, 5, 7...$;
3 - again, write the sequence of differences between consecutive terms of the previous sequence: $2, 2, 2, 2...$.
Because the terms of this third sequence are constant, the sum is a cubic of the form
Since the sum is zero for $k=0$, it follows that $d=0$. Then, we can identify coefficients rewriting the equation $ak^3+bk^2+ck$ with $k=1$, $k=2$, and $k=3$ (for which we know the corresponding results $1, 5, 14$) obtaining the following three equations:
The, we can solve the system, getting $a=1/3$, $b=1/2$, and $c=1/6$. This leads to
$$1/6\,(2k^3+3k+k)=1/6 \, k(k+1)(2k+1)$$