Proving formula for sum of squares with binomial coefficient $$\sum_{k=0}^{n-1}(k^2)= \binom{n}{3} + \binom{n+1}{3}$$
How should I prove that it is the correct formula for sum of squares?
Should I use induction to prove the basis? Any help is appreciated.

Using repeated differences we get
$$
\begin{array}{llll}
0 & 1 & 5 & 14 & \\
1 & 4 & 9 & \\
3 & 5 & \\
2& \\
\end{array}
$$
The last difference repeats forever, because $k^2$ has degree $2$.
Newton's interpolation formula then gives us this formula for the sequence of sums:
$$
0 \binom{n-1}{0} + 1 \binom{n-1}{1} + 3 \binom{n-1}{2} + 2 \binom{n-1}{3}
$$
Using Pascal's rule repeatedly we get
$$
\binom{n-1}{1} + \binom{n-1}{2} + \binom{n-1}{2} +  \binom{n-1}{2} + \binom{n-1}{3} + \binom{n-1}{3}
$$
$$
=\binom{n}{2} + \binom{n-1}{2} + \binom{n}{3} + \binom{n-1}{3}
$$
$$
=\binom{n}{2} + \binom{n}{3} + \binom{n-1}{2} + \binom{n-1}{3}
$$
$$
=\binom{n+1}{3} + \binom{n}{3}
$$
