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"If in the obvious equalities $(k+1)^3−k^3=3k^2+3k+1$, for the different values $k=1,2,…,n−1$, we add the left and the right sides separately, we obtain the equation $n^3−1=3σ_n+\frac{3(n−1)n}{2}+n−1$, where $σ_n=1^2+2^2+…+(n−1)^2$."

I'm stuck trying to understand what the author has done in the paragraph below. Is it possible to explain it using high school-level math? For instance, what is meant with "for the different values $k=1,2,…,n−1$, we add the left and the right sides separately"? I don't understand what he does after that either.

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up vote 3 down vote accepted

This equality


is true for every $k$, in particular it is true when $k=1$,$k=2$,etc... Writing all these equations, we get

$$\begin{array}{rcl} (1+1)^3−1^3&=&3\cdot 1^2+3\cdot 1+1\\ (2+1)^3−2^3&=&3\cdot 2^2+3\cdot 2+1\\ &\vdots &\\ ((n-2)+1)^3−(n-2)^3&=&3\cdot (n-2)^2+3\cdot (n-2)+1\\ ((n-1)+1)^3−(n-1)^3&=&3\cdot (n-1)^2+3\cdot (n-1)+1\\ \end{array}$$ Then you just sum all these equations (right hand sides with right hand sides, left hand sides with left hand sides) to find

$$\color{red}{2^3}-1^3+\color{blue}{3^3}-\color{red}{2^3}+4^3-\color{blue}{3^3}+\ldots +\color{green}{(n-1)^3}−(n-2)^3+n^3−\color{green}{(n-1)^3} = n^3-1$$

for the left hand side (note that this is called a telescopic sum). Now for the other side of the equation, note that $$3\cdot 1^2 + 3 \cdot 2^2+ \ldots +3\cdot (n-2)^2+3\cdot (n-1)^2 = 3 \sigma_n$$ where $σ_n=1^2+2^2+…+(n−1)^2.$ Furthermore $$3\cdot 1 + 3 \cdot 2+ \ldots +3\cdot (n-2)+3\cdot (n-1)= 3(1+2+\ldots +(n-1)) = 3\frac{n(n-1)}{2}. $$ (see the note for a proof of this fact) and $$\underbrace{1+ 1+\ldots +1 + 1}_{(n-1) \text{ times}} = n-1.$$ So you finally get $$n^3−1=3σ_n+\frac{3(n−1)n}{2}+n−1$$ NOTE: For any $k \in \mathbb{N},$ we have $$\begin{array}{rcl}2(1+2+\ldots+(k-1)+k) &=& (1+2+\ldots+k)+(k+(k-1)+\ldots+2+1)\\ &=& (1+k)+(2+(k-1))+\ldots +((k-1)+2)+(k+1)\\ &=& k(k+1)\end{array}$$

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Could you show me how I can sum those equations? I tried doing it myself, but I don't know what to do with the terms betweeen the first two and the last two terms. – Ali Mustafa Jul 6 '14 at 10:56
@AliMustafa I edited my answer and hope it answers better your question now. – Surb Jul 6 '14 at 11:23
How did you get from 1+2+…+(n−1) to n(n−1)2? – Ali Mustafa Jul 6 '14 at 11:54
@AliMustafa See my note for a proof of this fact... (with $k = n-1$) – Surb Jul 6 '14 at 11:55
@AliMustafa $(2n-1)(n-1) = 2n^2-2n-n+1 = 2n^2-3n+1$ – Surb Jul 6 '14 at 13:03

It's a concise way of explaining the following:

We wish to calculate $\displaystyle \sum_{i=1}^ni^2$ by considering $S_n =\displaystyle \sum_{i=1}^n i^3$, and we know (by evaluating) that $(k+1)^3 - k^3 = 3k^2 + 3k+ 1$

So $$S_n - S_{n-1}= 1+\sum_{i=1}^{n-1} (i+1)^3 - i^3 = \sum_{i=1}^{n-1} 3i^2+ 3i +1 = 3\sigma_n + \frac{3n(n-1)}2 + n$$

But equally, $S_n - S_{n-1} = n^3$ by cancelling terms:$$S_n - S_{n-1} = 1+\sum_{i=1}^{n-1} (i+1)^3 - i^3 = 1+n^3 - \left ( (n-1)^3 - (n-1)^3\right ) + \ldots + \left (2 - 2\right) - 1=n^3$$

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He performed a summation over the given value of k on each side. You may want to look into series to get a clearer idea.

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This is a finite sum, no series are required. – Jack M Jul 6 '14 at 10:26
A finite sum is a special case of series called finite series. @Jack M – Fermat Jul 6 '14 at 10:39
@Fermat True, but if you give the OP the impression that something called "series" are required to understand this, they'll google that, see the calculus stuff, and potentially get scared off (given their request to explain it using high-school math). – Jack M Jul 6 '14 at 11:27
@Jack M You are right. – Fermat Jul 6 '14 at 13:10

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