geometry - Method of Exhaustion applied to Parabolic Segment in Apostol's Calculus

In section I1.3 of Apostol's Calculus (2nd Ed., Vol. 1, pages 3 to 7), Apostol details how to apply the method of exhaustion to a the parabolic segment of $x^{2}$. I understand the process of applying the , however I'm not sure how he proceeds from a certain step to another. In particular, it is the step in which he introduces the inequality necessary to proceed to single out $b^3 \over 3$ as the value of the area of the parabolic segment. I've attempted to provide as much context as possible below so that reference to the book isn't necessary.

So in his explanation, he employs the method of exhaustion on the parabolic segment of $x^2$ from $x=0$ to $x=b$ with outer rectangles and inner rectangles, also known and upper and lower rectangles in other books.

The area of the lower rectangles is:

$$ S_{inner}= {b^3 \over n^3}[1^{2}+2^{2}+...+(n-1)^{2}] $$

$$ S_{outer}= {b^3 \over n^3}[1^{2}+2^{2}+...+n^{2}] $$

He then says that calculating the sum of the squares is inconvenient, and introduces the identity:

$$ (I.3) \ 1^{2}+2^{2}+...+n^{2} = {n^3 \over 3}+{n^2 \over 2}+{\frac n6}$$

and says that for the summation of squares from $1$ to $(n-1)$:

$$ (I.4) \ 1^{2}+2^{2}+...+(n-1)^{2} = {n^3 \over 3}-{n^2 \over 2}+{\frac n6}$$

He says, "For our purposes, we do not need the exact expressions given in the right-hand members of (I.3) and (I.4). All we need are the two inequalities:

$$1^{2}+2^{2}+...+(n-1)^{2} < {n^{3} \over 3} < 1^{2}+2^{2}+...+n^{2}$$

What I don't understand is where he obtained the idea to use ${n^{3} \over 3}$ in the inequality, from the expressions I've provided above. It seems to me that he used his prior knowledge of the value of the area to continue the proof, which leaves the reader wondering where he got the value from. He could have also used the following valid inequality:

$$1^{2}+2^{2}+...+(n-1)^{2} < {n^{3} \over 3}+ \frac n6 < 1^{2}+2^{2}+...+n^{2}$$

This is true if you simply remove the $n^2 \over 2$ term in I.3 and I.4 above. But he instead chose to use the first inequality. If there is a reason for why, I'd like to know.

Finally, and this is certainly pedantic on my part, I'd like to note that I think his proof of the area of a parabolic segment is inappropriately named the method of exhaustion. Isn't the reason Archimedes' method of analysis was heralded as ahead of its time is because he used the notion of the limiting process (albeit indirectly)? When I first encountered this method, the values of the inner and outer rectangles I provided above were subjected to the limiting process which results in the area under the parabolic segment. That seems more akin to the process with which Archimedes approached the problem (which was simply adding more and more triangles to fill in the empty space between the circle, and the polygon inscribed in the circle).