Prove that $1^2 + 2^2 + {...}+ {(n - 1)}^2 < \frac{n^3}{3} < 1^2 + 2^2 + {...}+ {n}^2 $ Prove that $1^2 + 2^2 + {...}+ {(n - 1)}^2 < \frac{n^3}{3} < 1^2 + 2^2 + {...}+ {n}^2 $

I know I need to use induction for this proof, but it feels like a pretty complicated one.
Basis: For $n = 2$,
$$1^2 < \frac{8}3 < 1^2 + 2^2$$
Induction Hypothesis: Assume $P(n)$ holds for $n=k$, that is,
$$1^2 + 2^2 + {...}+ {(k - 1)}^2 < \frac{k^3}{3} < 1^2 + 2^2 + {...}+ {k}^2$$
We need to show that $P(n)$ also holds for $n=k+1$
Proof:
$$1^2+2^2+{...}+{(k)}^2=1^2+2^2+{...}+{(k-1)}^2+{k}^2$$
After this, I'm not sure how to use the assumed inequality to prove it because it's a less than inequality. If I could get a hint that'd be awesome.

Since $x^2$ is an increasing function on $[0,1]$ we have that $\frac{1}{3}=\int_{0}^{1}x^2\,dx$ can be bounded by two Riemann sums:
$$ \frac{1}{n}\sum_{k=0}^{n-1}\left(\frac{k}{n}\right)^2 < \int_{0}^{1}x^2\,dx < \frac{1}{n}\sum_{k=1}^{n}\left(\frac{k}{n}\right)^2 $$
i.e. the lower/upper Riemann sums associated to the partition of $[0,1]$ into $n$ congruent sub-intervals.
By multiplying each term by $n^3$ the claim is readily proved.

Through induction we just have to prove that $\frac{1}{3}\left[(n+1)^3-n^3\right]$ is bounded between $n^2$ and $(n+1)^2$, i.e.
$ n^2<n^2+n+1<n^2+2n+1 $, which is also pretty straightforward.

Yet another way: $\sum_{m=0}^{M}m^2$ can be computed through the hockey stick identity. We have $m^2=2\binom{m}{2}+\binom{m}{1}$, hence
$$ \sum_{m=0}^{M}m^2=2\sum_{m=0}^{M}\binom{m}{2}+\sum_{m=0}^{M}\binom{m}{1} =2\binom{M+1}{3}+\binom{M+1}{2}=\frac{M(M+1)(2M+1)}{6}$$
and the claim is equivalent to
$$ M(M-1)(2M-1) < 2M^3 < M(M+1)(2M+1). $$
