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So I know of the basic summation:

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

You derive this by noting that if you pair every element with the one on the other end (example: $1$ with $n$, $2$ with $n - 1$) and so on, you get $\frac{n}{2}$ pairs of $n + 1$.

For

$$1^3 + 2^3 + ... + n^3,$$ one way to note that it is $$\left(\frac{n(n +1)}{2}\right)^2$$ is by observing values, but this is not good enough for me, since it does not build intuition, and let me see how I can "derive" it more formally (as with the above pairing for the original summation) myself. How can I "derive" it myself, through some sort of scheme, similar to the pairing scheme I described for the summation $1 + 2 + \dots + n$?

Blue
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user129393192
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2 Answers2

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We will use the concept of telescoping series:

We know that $(a-1)^4 = a^{4}-4a^{3}+6a^{2}-4a+1$

Let $a = n$

$=> (n-1)^4 = n^4 - 4n³ + 6n² - 4n + 1 \quad$

$=> \quad n^4 \quad- (n-1)^4 = 4n^3 - 6n^2 + 4n - 1 \qquad$ ...(1)

Since this is an identity, it is true for every $n$, it will also be true for $n-1$ (you can also understand it by letting $a = n-1$ from the original $(a-1)^4$ identity and again rearranging)

$=> (n-1)^4 - (n-2)^4 = 4 (n-1)^3 - 6(n-1)² + 4(n-1) - 1 \qquad$ ...(2) $=>(n-2)^4 - (n-3)^4 = 4 (n-2)^3 - 6(n-2)² + 4(n-2) - 1 \qquad$ ...(3)

We keep going down until we have

$=> \quad1^4 \quad-\quad 0^4 \quad = 4(1-0)³-6(1-0)² +4(1-0)-1 \qquad$ ...(n)

Adding all the equations (1) through (n), observe that every right term on the LHS (except $0^4$) is being cancelled out with every left term of the previous iteration (except $n^4$). Also we can simplify the RHS as well by combining the similar power terms. We result in the following

$=>n^4 - 0^4 = 4 \cdot (1^3+2^3+3^3...+n^3) - 6(1^2+2^2+3^2+...+n^2) + 4(1+2+3+...+n) - 1 \cdot n$

Substituting, $(1^2+2^2+3^2+...+n^2) = \frac{N \cdot (N+1) \cdot (2N+1)}{6} \quad$ and $ \quad (1+2+3+...+n) = \frac{N \cdot (N+1)}{2}$

You can simplify the equation to get $$(1^3+2^3+3^3...+n^3) = \left( \frac{n \cdot (n+1)}{2} \right)^2 $$

Generalisation: Using this approach you can inductively find any sum of ith powers from 1 to n by taking $n^{i+1} - (n-1)^{i+1}$ and following the same procedure as above

Devansh Agarwal
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  • You suddenly introduce a notation $S(k)$ which you do you define, and which seems to depend on $n,$ not just $k.$ – Thomas Andrews Oct 07 '23 at 23:15
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    This answer needs much more explanation. I do not understand it at all. Lots of steps seem to be skipped, especially with what f(n) is supposed to be. – user129393192 Oct 08 '23 at 02:04
  • @user129393192 I disregarded the notation f(n), S(i) as well as the parameter N to avoid any confusion. Made more than a few changes. I hope you understand the proof now. – Devansh Agarwal Oct 08 '23 at 05:34
  • I figured out what you meant, but readers shouldn't have to figure out what you mean, you should explicitly define it. Also, $N$ is not $n.$ @DevanshAgarwal – Thomas Andrews Oct 08 '23 at 15:24
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If you're comfortable with higher dimensions, you can think of the sum of cubes as the hypervolume of a 4-d discrete hyperpyramid. I put together an interactive a while back for visualizing this 4-d structure here: https://trkern.github.io/faulhaber.html

TomKern
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