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I'm having a difficulty understanding this enter image description here

It's only the addition part which I'm not following.

From what I've understood:

If we have a certain number of formulas and we add them, this is what happens on the left side:

$$3\cdot 1^2+3\cdot 2^2+\cdots+3\cdot (n-1)^2=3\cdot(1^2+2^2+\cdots+(n-1)^2)$$

$$3\cdot 1+3\cdot 2+\cdots +3\cdot (n-1)=3\cdot (1+2+\cdots+(n-1))$$

and we know that when we have a sequence of formulas, it will be $(n-1)$ formulas, so we represent the total numbers of $1$'s in the form $(n-1)$

Finally,

$$3(1^2+2^2+\cdots+(n-1)^2)+3(1+2+\cdots+(n-1))+(n-1)=\cdots$$

Now, for the right hand side, if we have, as an example:

$$2^3-1^3$$

$$3^3-2^3$$

$$4^4-3^3$$

$$n^3-4^3$$

We can reorganize this:

$$2^3-1^3$$

$$-2^3+3^3$$

$$4^4-3^3$$

$$n^3-4^3$$

We cancel which leaves us with

$$0-1^3$$

$$4^4+0$$

$$n^3-4^3$$

Rearrange and cancel:

$$0-1^3$$

$$0+4^4$$

$$n^3-4^3$$

Which leaves us with

$$n^3-1^3$$

Finally $$3\cdot(1^2+2^2+\cdots+(n-1)^2)+3\cdot(1+2+\cdots+(n-1))+(n-1)=n^3-1^3$$ The author doesn't say how exactly he did it, but would this way be the correct one? Or it's not the correct way?

Thank you!

Ian Mateus
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3 Answers3

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The idea of the proof is to use that $$\underbrace{1+1+\cdots+1+1}_n=n$$ $$1+2+\cdots+k=\dfrac{n(n+1)}2$$ and telescopy. We write $(k+1)^3-k^3=3k^2+3k+1$ using the binomial theorem. We can do this for $k=1,2,\ldots,n$. This gives $$\begin{align} (1+1)^3-1^3&=3\cdot \color{red}{1^2}+3\cdot \color{green}{1}+\color{blue}{1}\\ (2+1)^3-2^3&=3\cdot \color{red}{2^2}+3\cdot \color{green}{2}+\color{blue}{1}\\ \vdots\;\;\;\;\;\;\;\;\;&=\;\;\;\;\;\;\;\;\;\vdots\\ (n+1)^3-n^3&=3\cdot \color{red}{n^2}+3\cdot \color{green}{n}+\color{blue}{1} \end{align}$$ Summing all this gives that $$(n+1)^3-1^3=3\color{red}{(1^2+2^2+\cdots +n^2)}+3\color{green}{(1+2+\cdots+n)}+\color{blue}{(1+1+\cdots+1+1)}$$

because on the left side, $(1+1)^3$ cancels with $-2^3$, $(2+1)^3$ cancels with $-3^3$, and so on.

Pedro
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  • I don'T know calculus yet... he does it without calculus. Another way of explaining it ?? Thank you ! –  Dec 22 '13 at 22:04
  • @Astroman What do you mean by "calculus"? If you're not comfortable with $\Sigma$ notation, one can write the above replacing $\sum_{k=1}^n k^2$ with $1^2+2^2+\cdots+n^2$ and so on. – Pedro Dec 22 '13 at 22:06
  • Well, I meant that I have no idea what this greek symbol means, yet. –  Dec 22 '13 at 22:07
  • Right, I suspected so. Let me fix that. – Pedro Dec 22 '13 at 22:08
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    @Astroman: For what it's worth, a lot of beautiful mathematics is written using $\sum$ and $\prod$ (the analogous notation for products) and in hindsight I wish I had gotten comfortable with the notations sooner. Now that you know the idea, you might consider rereading Pedro's original post (click the link in the middle of the bottom of the post) to try to familiarize yourself with it :) – Eric Stucky Dec 22 '13 at 23:49
  • @EricStucky Agreed. – Pedro Dec 23 '13 at 00:05
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One way to show this is as follows: let's call the sum we seek $S$: $$ S = 1 + 4 + 9 + \ldots + n^2. $$ Notice that $$ \begin{align} 1 &= 1\\ 4 &= 1 + 3\\ 9 &= 1 + 3 + 5\\ \ldots \end{align} $$ In general, we have $$ n^2 = \sum_{k=1}^n(2k-1), $$ which follows immediately from the fact that $$ \sum_{k=1}^n k = \frac{n(n+1)}{2}. $$ Therefore, the total sum is equal to $$ S = \sum_{l=1}^n\sum_{k=1}^l(2k-1). $$ If we change the order of the sums, we find $$ S = \sum_{k=1}^n\sum_{l=k}^n(2k-1) = \sum_{k=1}^n(2k-1)\sum_{l=k}^n 1= \sum_{k=1}^n(2k-1)(n+1-k). $$ You can also see this from the fact that $$ \begin{align} 1 &= 1\\ 4 &= 1 + 3\\ 9 &= 1 + 3 + 5\\ \ldots\\ n^2 &= 1 + 3 + 5 + \ldots + 2n-1\\ \hline\\ S &= 1\cdot n + 3(n-1) + 5(n-2) + \ldots + (2n-1)\cdot 1\\ &= (2\cdot 1-1)(n+1-1) + (2\cdot 2-1)(n+1-2) + \ldots + (2n-1)(n+1-n)\\ &= \sum_{k=1}^n(2k-1)(n+1-k). \end{align} $$ If we work this out, we get $$ \begin{align} S &= (n+1)\sum_{k=1}^n(2k-1) - 2\sum_{k=1}^n k^2 + \sum_{k=1}^n k\\ &= (n+1)n^2 - 2S + \frac{n(n+1)}{2}. \end{align} $$ Therefore, $$ 3S = n^3 + \frac{3n^2}{2} + \frac{n}{2}, $$ so that $$ S = \frac{n^3}{3} + \frac{n^2}{2} + \frac{n}{6}. $$

Pulsar
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There is another simple way to do this. $$\begin{align}\sum_{k=1}^nk^2&=\sum_{k=1}^nk(k-1)+k\\&=\sum_{k=1}^n\frac{(k+1)k(k-1)-k(k-1)(k-2)}{3}+\sum_{k=1}^nk\\&=\frac{(n+1)n(n-1)}{3}+\frac{n(n+1)}{2}\\&=\frac{1}{3}n^3+\frac{1}{2}n^2+\frac{1}{6}n\end{align}$$

Richard
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