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.

It's easy to prove this directly from the definition of the binomial coefficient. The advantage is you don't need to remember any binomial identities, or try to rewrite a fraction to look like a binomial.
The left-hand side is $$\sum_{k=0}^{n-1}k^2=\frac{n(n-1)(2n - 1)}{6}$$ by a well known formula (you can also prove this by induction very easily).
The right-hand side is $$\binom{n}{3} + \binom{n + 1}{3}=\frac{n!}{3!(n - 3)!} + \frac{(n + 1)!}{3!(n + 1 - 3)!} = \frac{n! \left(n - 2\right)! + \left(n - 3\right)! \left(n + 1\right)!}{6 \left(n - 3\right)! \left(n - 2\right)!}$$
Writing $n! = n(n - 1)(n - 2)(n - 3)!$ we can cancel $(n - 3)!$ giving $$\frac{n(n - 1)(n - 2) \left(n - 2\right)! +  \left(n + 1\right)!}{6 \left(n - 2\right)!}.$$ Similarly we can write $(n + 1)! = (n + 1)n(n - 1)(n - 2)!$ and cancel $(n -2)!$ giving $$\frac{n(n - 1)(n - 2) +  \left(n + 1\right)n(n - 1)}{6}.$$ Factoring $n(n - 1)$ in the numerator gives $$\frac{n(n - 1)(n - 2 + n + 1)}{6} = \frac{n(n - 1)(2n - 1)}{6}.$$
