Prove by induction that $ 1^2+2^2+...+(n-1)^2In Apostol's «Calculus I» on page 33 there is the following proof by induction:

To prove:
  $$
1^2+2^2+...+(n-1)^2<n^3/3<1^2+2^2+...+n^2
$$
  Solution:
  Consider the leftmost iequality first, and left us refer to this formula as $A(n)$. It is easy to verify this assertion directly for the first few values of n. Thus, for example, when n takes the values 1, 2, 3, the assertion is true.
  Our object is to prove that $A(n)$ is true for every positive integer n. The procedure is as follows: Assume the assertion has been proved for a particular value of n, say for $n = k$. That is, assume we have proved
  $$
A(k): 1^2+2^2+…+(k-1)^2<\frac{k^3}{3}
$$
  for a fixed $k\geqslant 1$. Now using this, we shall deduce the corresponding result for $k+1$:
  $$
A(k+1): 1^2+2^2+…+k^2<\frac{(k+1)^3}{3}
$$
  Start with $A(k)$ and add $k^2$ to both sides. This gives the inequality
  $$
1^2+2^2+…+k^2<\frac{k^3}{3}+k^2
$$
  To obtain $A(k+1)$ as a consequence of this, it suffices to show that
  $$
\frac{k^3}{3}+k^2<\frac{(k+1)^3}{3}
$$

So my question is why does it suffice to to show the last inequality is true to finish the proof by induction? I don't see how is $A(k+1)$ follows from $A(k)$. It is clear to me that
$$
A(k) + k^2 < A(k+1)
$$
But it is unclear why does it make sense in this case.
 A: If $a<b$ and $b\le c$, then $a<c$.
A: You know $A(k)$:$$1^2+2^2+\dots+(k-1)^2 < k^3/3$$.
Add $k^2$ to both sides and you know:
$$1^2+2^2+\dots+k^2 < k^3/3 + k^2$$
If you know that $k^3/3+k^2 < (k+1)^3/3$, the you know $A(k+1)$ since:
$$1^2+2^2+\dots+k^2  < (k+1)^3/3$$
A: \begin{align*}
1^2+2^2+.....+(k-1)^2+k^2 &<\frac{k^3}{3}+k^2(\because A(k)\ \  \text{is true})\\
&=\frac{k^3+3k^2+3k+1}{3}-\frac{3k+1}{3}\\
&<\frac{(k+1)^3}{3}
\end{align*}
Therefore the result is true for A($k+1$) if A($k$) is true.
A: Probably easier but a bit too straightforward and slow. For the left inequality add $n^2$ on both sides. LHS becomes
$$
LHS=\frac{n^3}{3}+\frac{n^2}{2}+\frac{n}{6}
$$
and RHS is 
$$
RHS=\frac{n^3}{3}+n^2
$$
After all cancellations you get $LHS=n, \ RHS=3n^2$. For the second inequality, add $(n+1)^2$ on both sides. On the RHS after a bit of algebra you get
$$
\frac{n^3}{6}+\frac{n^2}{2}+\frac{n}{6}+n^2+2n+1
$$
After all cancellations you get $0$ on LHS and $\frac{n^2}{6}+\frac{n}{6}$ on RHS, hence both inequalities are proven. Whether this is strict inductive proof, I'm not too sure, more like perturbation method in Concrete Mathematics
