Hello, I'm trying to understand proof of Archimedes parabolic area of segment wrapped in modern notations but i'm little confused ( i mean confused as hell), i'm gonna throw a lot of questions now most of them stupid, so yeah enjoy,
1) The book says that when $\\n$ gets larger in equation (I.2) the computation of the sum would be very tedious but happily there's an identity that would simplify the things, that is equation (I.3), so it starts to prove that latter, and the first step is "let's consider the equation $\\(k+1)^3 = k^3 + 3k^2 + 3k + 1$ " and here the first question arises why would we starts with the proposition $\\(k+1)^3 = k^3 + 3k^2 + 3k + 1$ to prove our some's (I.1) and (I.2) identity , which in my eyes, i see no relation between that proposition and our sums, why ?
2) And it goes by saying as $\\k = 1,2,…(n-1)$ we could rewrite the equation as $\\3(n-1)^2+3(n-1)+1 = n^3-(n-1)^3$ and "When we add these formulas, all the terms on the right cancel except two and we get $\\3[1^2+2^2….+(n-1)^2] + 3[1+2…+(n-1)]+(n-1) = n^3-1^3$" ahaa, how this $\\3(n-1)^2+3(n-1)+1 = n^3-(n-1)^3$ can lead to this $\\3[1^2+2^2….+(n-1)^2] + 3[1+2…+(n-1)]+(n-1) = n^3-1^3$ i tried to compute it my self many times but i didn't get the same result, so how?
3) And it goes by saying that the second sum on the left is an arithmetic progression and it could simplify to $\frac{1}{2}n(n-1)$, how , i tried to compute $\frac{1}{2}n(n-1)$ by substituting $n$ with arithmetic sequence($\frac{1}{2}0(0-1)$,$\frac{1}{2}1(1-1)$….) and it gave me the result ($0,1, 3, 6, 10, 15, 21, 28….$) and for the second sum in the left trying to do it the same way would give (31=3, 32 = 6, 9, 12, 15, 18…) , definitely not the same results, i know i'm wrong, tell me where and why ?
4) Afterwards it says that the second sum in the left could be simplified to $\frac{1}{2}n(n-1)$ and so we can deduce the equation (I.4) again confused on how the writer have gone from $\\3[1^2+2^2….+(n-1)^2] + 3[1+2…+(n-1)]+(n-1) = n^3-1^3$ to $1^2+2^2+…+(n-1)^2 = \frac{n^3}{3} – \frac{n^2}{2} + \frac{n}{6} ? $
5) The writer continue by saying "by adding $n^2$ to both members we get our proof (I.3) " and again confused how adding $n^2$ to both side could lead us to the proof expression ?
ok , sorry for the long list, but i'm sure a lot of people (beginners like me) is having the same questions while reading the famous book "Calculus, Vol. 1 One-Variable Calculus, with an Introduction to Linear Algebra by Tom M. Apostol " and i thought why not put a detailed questioning post to clear confusion for newcomers , hope you have the patient to read through this, feel free to answer how many questions you like and to be as detailed and clear as possible thanks a lot.
Best Answer
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.