Simplifying a sum of consecutive cubes

If you were asked to find the sum of the first four cubes, you would easily be able to do so:

$1^3 + 2^3 + 3^3 + 4^3 = 100$

But here you may notice something interesting:

$1^3 + 2^3 + 3^3 + 4^3 = (1+2+3+4)^2$

It turns out that this holds for the sum of the first $n$ cubes for any $n$ — an identity. This identity can be written also as:

$\displaystyle \sum_{i=1}^n i^3 = \left( \displaystyle \sum_{i=1}^n i \right)^2$

A second way of expressing the same identity simplifies the right side and expresses it as a combination:

$\displaystyle \sum_{i=1}^n i^3 = \binom{n+1}{2}^2$

A proof by bijection

This identity can be proven combinatorially by noticing that $\binom{n+1}{2}$ is the number of ways to choose 2 elements with repetition from the set {1..n} — the following is a proof by Benjamin and Orrison.

In this proof, we prove the third form of the identity

$\displaystyle \sum_{i=1}^n i^3 = \binom{n+1}{2}^2$

by constructing one set of size $\displaystyle \sum_{i=1}^n i^3$ and constructing another set of size $\binom{n+1}{2}^2$ , then constructing a bijection between the two sets.

Let $S$ be the set of ordered 4-pairs $(a,b,c,d)$ such that $1 \leq a,b,c \leq d \leq n$ — that is, the fourth integer of the pair is greater or equal to each of the other three. If we fix $d$ , the number of possibilities for the variables $a,b,c$ is $d^3$ . As $d$ ranges from 1 to n, the total number of elements in $S$ is equal to $\displaystyle \sum_{i=1}^n i^3$ .

Also, since $\binom{n+1}{2}$ is the number of ways to choose 2 elements with repetition from the set {1..n}, there are $\binom{n+1}{2}$ pairs of integers $(h,i)$ satisfying $1 \leq h \leq i \leq n$ . So let $T$ be the set of pairs of such pairs $((h,i),(j,k))$ where $1 \leq h \leq i \leq n$ and $1 \leq j \leq k \leq n$ . The number of elements in $T$ is then $\binom{n+1}{2}^2$ .

We further partition the sets as follows: let $S_1$ be the subset of $S$ where $a \leq b$ and let $S_2$ be the subset of $S$ where $a > b$ . Similarly, let $T_1$ be the subset of $T$ where $i \leq k$ and let $T_2$ be the subset of $T$ where $i > k$ . We can construct a bijection between $S$ and $T$ by constructing two bijections:

$S_1 \leftrightarrow T_1$

$S_2 \leftrightarrow T_2$

The following is trivially a bijection from $S_1$ to $T_1$ — the case where $a \leq b$ :

$(a,b,c,d) \rightarrow ((a,b),(c,d))$

The equivalent bijection from $S_2$ to $T_2$ — the case where $a > b$ :

$(a,b,c,d) \rightarrow ((c,d),(b,a-1))$

We can confirm that the two operations we defined are indeed bijections. This proves the identity.

3 thoughts on “Simplifying a sum of consecutive cubes”

JL says:

May 21, 2011 at 10:11 pm Reply

Would it not be easier to inductively prove that the sum of the first n integers is n(n+1)/2 and that the sum of the first n cubes is n^2(n+1)^2/4, which equals (n(n+1)/2)^2?

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1. luckytoilet says:
  
  May 21, 2011 at 10:13 pm Reply
  
  Shhhh 😛
  Besides, any bijective proof beats any boring inductive argument xD
  
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Anonymous says:

November 30, 2011 at 11:12 pm Reply

Bijections are awesome 😛
Your blog is so interesting

A Western Li

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Simplifying a sum of consecutive cubes

A proof by bijection

3 thoughts on “Simplifying a sum of consecutive cubes”

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A proof by bijection

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