I am trying to find a way to move from pre-calculus to learning calculus by reading Tom Apostol Calculus Volume One. Since this is the first time I am opening up a calculus book I am struggling to grasp the topic.
In Tom Apostol's book he is talking about Archimede's Method of Exhaustion.
Tom Apostol shows a bunch of steps on how Archimedes and other
mathematicians may have contributed to the idea of finding the area
of a shape that is unknown or hard to find. I was looking for some help on one of his steps.
I found this page on Math Stack Exchange but was looking to see if I understand
what Tom Apostol is talking about. Here is the link: YouTube Video showing how to cancel equations
The step I have questions on starts with an identity
$$(K+1)^3 = K^3+3K^2+3K+1$$ Equation is rewritten as $$3K^2+3K+1=(K+1)^3-K^3$$ I am good up to here, Tom Apostol subtracted the $K^3$ from both sides. The next step then is to start plugging in integers. The identity up to this point should be satisfied for every integer $n\ge 1$. Tom Apostol sets up this step like this: $$\begin{array}{L}3* 1^2 + 3 * 1 + 1 = 2^3 -1^3 \\3 * 2^2 +3 * 2 + 1 = 3^3-2^3\\ \phantom2 \vdots \\ 3(n-1)^2+3(n-1)+1 = n^3-(n-1)^3\end{array}$$
I have an idea of what Tom Apostol comes up with the right side of the equation.
$$\begin{array}{L} = \phantom9 \phantom1 \require{cancel}\cancel{2^3}- 1^3 \\ = -( \cancel{3^3}-\cancel{2^3} )\\ = -( 4^3 – \cancel{3^3}) \\ = – (n-1)^3 + 1^3 \end{array}$$ Am I going in the right direction? I feel like I'm close to understanding the right side, but can see my method is not exact if it is in the right direction. Thanks for any help.
Best Answer
What Tom Apostol want to do in that part of the book is to show that for any $n\geq 1$ you have: $$ 1^2+2^2\dots+n^2=\frac{n^3}{3} + \frac{n^2}{2}+\frac{n}{6} $$ As you pointed out, he starts with the identity $$ 3k^2+3k+1 = (k+1)^3-k^3 $$ from which the following equations follow, and proceeds to add all them up: $$ \begin{array}{lrl} &3\cdot 1^2+3\cdot 1+1&=2^3-1^3 \\ &3\cdot 2^2+3\cdot 2+1&=3^3-2^3 \\ &\vdots & \\ +&3(n-1)^2+3(n-1)+1&=n^3-(n-1)^3\\ \hline\\ &3[1^2+2^2+\dots+(n-1)^2]+3[1+2+\dots+(n-1)]+(n-1) &=n^3-1^3 \end{array} $$ Let's see how he got all the terms one by one. First, the left-hand side is composed of three terms: $3[1^2+2^2+\dots+(n-1)^2]$ plus $3[1+2+\dots+n]$ plus $(n-1)$. These terms are obtained by adding the first, second and third terms of the right hand side of all $(n-1)$ equations independently: $$ \begin{aligned} 3\cdot 1^2 + 3\cdot 2^2+\cdots+3\cdot(n-1)^2 &= 3[1^2+2^2+\cdots+(n-1)^2]\\ 3\cdot 1 + 3\cdot 2+\cdots+3\cdot(n-1) &= 3[1+2+\cdots+(n-1)]\\ 1+1+\cdots+1&=(n-1) \end{aligned} $$ Now, lets look at the right hand side. Similarly we add the first and second terms of the $(n-1)$ by separate and then add them at the end: $$ \begin{array}{cccccccc} & &+2^3&+3^3&+\cdots&+(n-1)^3&+n^3\\ +&-1^3&-2^3&-3^3&-\cdots&-(n-1)^3\\ \hline\\ &-1^3+&+0&+0&+\cdots&+0&+n^3 \end{array} $$ The next step is to substitute the well known formula for $1+2+\dots+(n-1) = \frac{n(n-1)}{2}$ to obtain: $$ 3[1^2+2^2+\dots+(n-1)^2]+\frac{3n(n-1)}{2}+(n-1) =n^3-1^3 $$ and rearrange to obtain: $$ \begin{aligned} 1^2+2^2+\dots+(n-1)^2 &= \frac{1}{3}\left(n^3-1^3 - \frac{3n(n-1)}{2}-(n-1)\right)\\ & = \frac{n^3}{3} -\frac{n^2}{2} + \frac{n}{6} \end{aligned} $$ Finally, just add $n^2$ to both sides of the previous equation to obtain the desired result: $$ \begin{aligned} 1^2+2^2+\dots+(n-1)^2 +n^2&= \frac{n^3}{3} -\frac{n^2}{2} + \frac{n}{6}+n^2\\ &=\frac{n^3}{3} + \frac{n^2}{2}+\frac{n}{6} \end{aligned} $$ Hopefully this will help you in addition to what Tom Apostol already explains in the book. Good luck!