Prove that $1^2 + 2^2 + ..... + (n-1)^2 < \frac {n^3} { 3} < 1^2 + 2^2 + ...... + n^2$ I'm having trouble on starting this induction problem.
The question simply reads : prove the following using induction: 
$$1^{2} + 2^{2} + ...... + (n-1)^{2} < \frac{n^3}{3} < 1^{2} + 2^{2} + ...... + n^{2}$$

I would start by proving each side of the equation using induction. 
First prove that $1^2+2^2+..+n^2 = \sum\limits_{i=1}^n i^2  \geq \frac{n^3}{3}$
Induction says that we must first check that the relationship holds for $n=1$.
$\sum\limits_{i=1}^1 i^2 = 1 \geq \frac{1^3}{3}=\frac{1}{3}$
So it holds for $n=1$. Now we assume that it is true that $\sum\limits_{i=1}^n i^2  \geq \frac{n^3}{3}$ for $n=k$, and prove that the result can be obtained from $n=k+1$.
$\sum\limits_{i=1}^{k+1} i^2  = \sum\limits_{i=1}^k i^2  + (k+1)^2$
$\sum\limits_{i=1}^{k+1} i^2  \geq \frac{k^3}{3}  + k^2+2k+1$
$\sum\limits_{i=1}^{k+1} i^2  \geq \frac{k^3+3k^2+3k+1}{3}+k+\frac{2}{3}$
$\sum\limits_{i=1}^{k+1} i^2  \geq \frac{(k+1)^3}{3}+k+\frac{2}{3}$
$\sum\limits_{i=1}^{k+1} i^2  \geq \frac{(k+1)^3}{3}$
Since $k+\frac{2}{3}$ is a positive number, removing it from the right side still maintains the inequality, and the first part of your question has been proved by induction. Do a similar thing to the left hand side to prove it.
