A Detailed explanation for Archimedes area of parabolic segment by exhaustion proof ? Needed

I’ll deal first with the computational questions.

2) The single equation

$$3(n-1)^2+3(n-1)+1=n^3-(n-1)^3\tag{1}$$

doesn’t lead to the equation

$$\begin{align*} &3[1^2+\ldots+(n-1)^2]+3[1+\ldots+(n-1)]+(n-1)\\ &\qquad=n^3-1^3\,;\tag{2} \end{align*}$$

as the author says, $(2)$ is the result of adding up the $n-1$ equations in the list that ends with $(1)$. There is one of these for each value of $k$ from $1$ through $n-1$:

$$\begin{align*} \color{red}{3\cdot1^2}+\color{blue}{3\cdot1}+\color{green}1&=2^3-1^3\\ \color{red}{3\cdot2^2}+\color{blue}{3\cdot2}+\color{green}1&=3^3-2^3\\ &\;\;\vdots\\ \color{red}{3(n-1)^2}+\color{blue}{3(n-1)}+\color{green}1&=n^3-(n-1)^3 \end{align*}\tag{3}$$

On the left the red terms sum to

$$3[1^2+2^2+\ldots+(n-1)^2]\,,\tag{4}$$

the blue terms sum to

$$3[1+2+\ldots+(n-1)]\,,\tag{5}$$

and the green terms sum to $n-1$, giving us the lefthand side of $(2)$. The black terms on the righthand sides of $(3)$ sum to the total of these $n-1$ rows:

$$\begin{array}{ccc} \color{red}{2^3}&-&1^3\\ \color{blue}{3^3}&-&\color{red}{2^3}\\ \color{green}{4^3}&-&\color{blue}{3^3}\\ \color{brown}{5^3}&-&\color{green}{4^3}\\ &\vdots\\ \color{purple}{(n-1)^3}&-&\color{gray}{(n-2)^3}\\ n^3&-&\color{purple}{(n-1)^3} \end{array}$$

Each term in the first column except $n^3$, the last one, is matched by a corresponding negative term in the row below it in the second column: $\color{red}{2^3}$ and $-\color{red}{2^3}$, $\color{blue}{3^3}$ and $-\color{blue}{3^3}$, and so on. All of these colored terms cancel out, leaving only the black $n^3$ at the bottom of the first column and the black $-1^3$ at the top of the second column. It’s a telescoping sum

$$-1^3+\color{red}{2^3-2^3+3^3-3^3+\ldots+(n-1)^3-(n-1)^3}+n^3$$

in which all of the red terms cancel out, leaving only $n^3-1^3$.

3) The second sum on the left is $(5$) above. Ignore the factor of $3$ for a moment; the part inside the square brackets is simply the arithmetic progression

$$1+2+3+\ldots+(n-1)\,,$$

which has first term $1$, last term $n-1$, $n-1$ terms, so its sum is

$$\frac{(n-1)\big(1+(n-1)\big)}2=\frac{n(n-1)}2$$

by the standard formula for the sum of an arithmetic progression. You should know this formula, and you should also learn the special case $\sum_{k=1}^mk=\frac{m(m+1)}2$ for the sum of the first $m$ consecutive positive integers, since it comes up rather often.

4) At this point we have reduced $(3)$ to the equation

$$3[1^2+\ldots+(n-1)^2]+\color{red}{3\cdot\frac{n(n-1)}2+(n-1)}=n^3-1^3\,.$$

Transpose the red terms to the righthand side:

$$\begin{align*} 3[1^2+\ldots+(n-1)^2]&=n^3-1-3\cdot\frac{n(n-1)}2-(n-1)\\ &=n^3-1-3\cdot\frac{n(n-1)}2-n+1\\ &=n^3-3\cdot\frac{n(n-1)}2-n\,. \end{align*}$$

Divide through by $3$ and simplify:

$$\begin{align*} 1^2+\ldots+(n-1)^2&=\frac{n^3}3-\frac{n(n-1)}2-\frac{n}3\\ &=\frac{n^3}3-\frac{n^2-n}2-\frac{n}3\\ &=\frac{n^3}3-\frac{n^2}2+\frac{n}2-\frac{n}3\\ &=\frac{n^3}3-\frac{n^2}2+\frac{n}6\,. \end{align*}$$

5) We now have

$$1^2+\ldots+(n-1)^2=\frac{n^3}3-\frac{n^2}2+\frac{n}6\,,$$

and adding $n^2$ to both sides and simplifying gives us

$$\begin{align*} 1^2+\ldots+(n-1)^2+n^2&=\frac{n^3}3-\frac{n^2}2+\frac{n}6+n^2\\ &=\frac{n^3}3+\frac{n^2}2+\frac{n}6\,, \end{align*}$$

exactly as the author said.

Now back to your first question.

1) The short answer is because it works: we have a valid derivation of formula $(\text{I}.3)$. A slightly longer answer is that at some point someone was clever enough to realize by rewriting the simple power

$$(k+1)^3=k^3+3k^2+3k+1$$

as

$$(k+1)^3-k^3=3k^2+3k+1\,,$$

we can sum a bunch of similar equations (i.e., those in $(3)$ above) to express $n^3-1$ in terms of the sum of the first $n-1$ squares, the sum of the first $n-1$ positive integers, and $n$. We know the sum of the first $n-1$ positive integers as a function of $n$, so after a little rearrangement we find that we’ve expressed the sum of the first $n-1$ squares as a cubic polynomial in $n$, thereby getting a formula for the sum of the first $n$ squares.

The author probably doesn’t really expect you to see how someone would come up with this idea; he simply wants to present you with an elementary derivation of the formula, rather than give you the formula and ask you to take it on faith. There are actually quite a few ways to derive the formula, but this is perhaps the most elementary, as it requires only basic algebra.