Sum of Sequence of Cubes
Theorem
- $\ds \sum_{i \mathop = 1}^n i^3 = \paren {\sum_{i \mathop = 1}^n i}^2 = \frac {n^2 \paren {n + 1}^2} 4$
Proof by Induction
First, from Closed Form for Triangular Numbers:
- $\ds \sum_{i \mathop = 1}^n i = \frac {n \paren {n + 1} } 2$
So:
- $\ds \paren {\sum_{i \mathop = 1}^n i}^2 = \dfrac {n^2 \paren {n + 1}^2} 4$
Next we use induction on $n$ to show that:
- $\ds \sum_{i \mathop = 1}^n i^3 = \dfrac {n^2 \paren {n + 1}^2} 4$
The proof proceeds by induction.
For all $n \in \Z_{>0}$, let $\map P n$ be the proposition:
- $\ds \sum_{i \mathop = 1}^n i^3 = \dfrac {n^2 \paren {n + 1}^2} 4$
Basis for the Induction
$\map P 1$ is the case:
- $1^3 = \dfrac {1 \paren {1 + 1}^2} 4$
\(\ds \sum_{i \mathop = 1}^1 i^3\) | \(=\) | \(\ds 1^3\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {1^2 \paren {1 + 1}^2} 4\) |
Thus $\map P 1$ is seen to hold.
This is the basis for the induction.
Induction Hypothesis
Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
- $\ds \sum_{i \mathop = 1}^k i^3 = \dfrac {k^2 \paren {k + 1}^2} 4$
from which it is to be shown that:
- $\ds \sum_{i \mathop = 1}^{k + 1} i^3 = \dfrac {\paren {k + 1}^2 \paren {k + 2}^2} 4$
Induction Step
This is the induction step:
\(\ds \sum_{i \mathop = 1}^{k + 1} i^3\) | \(=\) | \(\ds \sum_{i \mathop = 1}^k i^3 + \paren {k + 1}^3\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {k^2 \paren {k + 1}^2} 4 + \paren {k + 1}^3\) | Induction Hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {k^4 + 2 k^3 + k^2} 4 + \frac {4 k^3 + 12 k^2 + 12 k + 4} 4\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {k^4 + 6 k^3 + 13 k^2 + 12 k + 4} 4\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\paren {k + 1}^2 \paren {k + 2}^2} 4\) |
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
- $\forall n \in \Z_{>0}: \ds \sum_{i \mathop = 1}^n i^3 = \dfrac {n^2 \paren {n + 1}^2} 4$
Proof by Nicomachus
By Nicomachus's Theorem, we have:
- $\forall n \in \N_{>0}: n^3 = \paren {n^2 - n + 1} + \paren {n^2 - n + 3} + \ldots + \paren {n^2 + n - 1}$
Also by Nicomachus's Theorem, we have that the first term for $\paren {n + 1}^3$ is $2$ greater than the last term for $n^3$.
So if we add them all up together, we get:
\(\ds \sum_{i \mathop = 1}^n i^3\) | \(=\) | \(\ds \sum_{\stackrel {1 \mathop \le i \mathop \le n^2 \mathop + n \mathop - 1} {i \text { odd} } } i\) | ... the sum of all the odd numbers up to $n^2 + n - 1$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{i \mathop = 1}^{\frac {n^2 \mathop + n} 2} 2 i - 1\) | ... that is, the first $\dfrac {n^2 + n} 2$ odd numbers | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\frac {n^2 + n} 2}^2\) | Odd Number Theorem |
Hence the result.
$\blacksquare$
Proof by Recursion
From Closed Form for Triangular Numbers:
- $(1): \quad \ds \map A n := \sum_{i \mathop = 1}^n i = \frac{n \paren {n + 1} } 2$
From Sum of Sequence of Squares:
- $(2): \quad \ds \map B n := \sum_{i \mathop = 1}^n i^2 = \frac{n \paren {n + 1} \paren {2 n + 1} } 6$
Let $\ds \map S n = \sum_{i \mathop = 1}^n i^3$.
Then:
\(\ds \map S n\) | \(=\) | \(\ds n^3 + \paren {n - 1}^3 + \paren {n - 2}^3 + \cdots + 2^3 + 1^3\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 0}^{n \mathop - 1} \paren {n - k}^3\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 0}^{n \mathop - 1} \paren {n^3 - 3 n^2 k + 3 n k^2 - k^3}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds n^4 - 3 n^2 \cdot \map A {n - 1} + 3 n \cdot \map B {n - 1} - \map S {n - 1}\) | ||||||||||||
\(\text {(3)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds \map S n\) | \(=\) | \(\ds n^4 - 3 n^2 \cdot \frac {n \paren {n - 1} } 2 + 3n \cdot \frac{n \paren {n - 1} \paren {2 n - 1} } 6 - \map S {n - 1}\) | substituting from $(1)$ and $(2)$ | |||||||||
\(\text {(4)}: \quad\) | \(\ds \map S n\) | \(=\) | \(\ds n^3 + \map S {n - 1}\) | recursive definition | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds 2 \map S n\) | \(=\) | \(\ds n^4 + n^3 - 3 n^2 \cdot \frac {n \paren {n - 1} } 2 + 3 n \cdot \frac {n \paren {n - 1} \paren {2 n - 1} } 6\) | adding $(3)$ and $(4)$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \) | \(=\) | \(\ds \frac {n^2 \paren {n + 1}^2} 2\) | simplification | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map S n\) | \(=\) | \(\ds \frac {n^2 \paren {n + 1}^2} 4\) |
$\blacksquare$
Proof using Bernoulli Numbers
From Faulhaber's Formula:
\(\ds \sum_{i \mathop = 1}^n i^p\) | \(=\) | \(\ds 1^p + 2^p + \cdots + n^p\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {n^{p + 1} } {p + 1} + \sum_{k \mathop = 1}^p \frac {B_k \, p^{\underline {k - 1} } \, n^{p - k + 1} } {k!}\) |
where $B_k$ are the Bernoulli numbers.
Setting $p = 3$:
\(\ds \sum_{i \mathop = 1}^n i^3\) | \(=\) | \(\ds \frac {n^{3 + 1} } {3 + 1} + \sum_{k \mathop = 1}^3 \frac {B_k \, 3^{\underline {k - 1} } \, n^{3 - k + 1} } {k!}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {n^4} 4 + \frac {B_1 \, 3^{\underline 0} \, n^3} {1!} + \frac {B_2 \, 3^{\underline 1} \, n^2} {2!} + \frac {B_3 \, 3^{\underline 2} \, n^1} {3!}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {n^4} 4 + \frac 1 2 \frac {n^3} {1!} + \frac 1 6 \frac {3 n^2} {2!} + 0 \frac {3 \times 2 n} {3!}\) | Definition of Bernoulli Numbers and Definition of Falling Factorial | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {n^4} 4 + \dfrac {n^3} 2 + \frac {n^2} 4\) | simplifying | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {n^2} 4 \paren {n^2 + 2 n + 1}\) | factor out $\dfrac {n^2} 4$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {n^2 \paren {n + 1}^2} 4\) |
Hence the result.
$\blacksquare$
Proof using Multiplication Table
- $\ds \paren {\sum_{i \mathop = 1}^n i}^2 = \paren {1 + 2 + 3 + \cdots + N}^2$
\(\ds \paren {1 + 2 + 3 + \cdots + N}^2\) | \(=\) | \(\ds 1 \times \paren {1 + 2 + 3 + \cdots + N}\) | ||||||||||||
\(\ds \) | \(+\) | \(\ds 2 \times \paren {1 + 2 + 3 + \cdots + N}\) | ||||||||||||
\(\ds \) | \(+\) | \(\ds \cdots\) | ||||||||||||
\(\ds \) | \(+\) | \(\ds N \times \paren {1 + 2 + 3 + \cdots + N}\) |
Organizing the terms above in a square matrix:
- $\begin{array}{r|cccccccccc} \paren {\ds \sum_{i \mathop = 1}^n i}^2 & 1 & 2 & 3 & 4 & 5 & 6 & \cdots & N \\ \hline 1 & 1 & 2 & 3 & 4 & 5 & 6 & \cdots & N \\ 2 & 2 & 4 & 6 & 8 & 10 & 12 & \cdots & 2 N \\ 3 & 3 & 6 & 9 & 12 & 15 & 18 & \cdots & 3 N \\ 4 & 4 & 8 & 12 & 16 & 20 & 24 & \cdots & 4 N \\ 5 & 5 & 10 & 15 & 20 & 25 & 30 & \cdots & 5 N \\ 6 & 6 & 12 & 18 & 24 & 30 & 36 & \cdots & 6 N \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ N & N & 2 N & 3 N & 4 N & 5 N & 6 N & \cdots & N^2 \\ \end{array}$
From Closed Form for Triangular Numbers:
- $\ds \sum_{i \mathop = 1}^n i = \frac {n \paren {n + 1} } 2$
Therefore:
- $\ds \paren {\sum_{i \mathop = 1}^n i}^2 = \dfrac {n^2 \paren {n + 1}^2} 4$
Next we notice that the sum of the terms in row $N$ with the terms in column $N$ (less $N^2$ so as not to double count) is precisely $N^3$.
\(\ds 1^3\) | \(=\) | \(\ds 1\) | ||||||||||||
\(\ds 2^3\) | \(=\) | \(\ds 2 + 4 + 2\) | ||||||||||||
\(\ds 3^3\) | \(=\) | \(\ds 3 + 6 + 9 + 6 + 3\) | ||||||||||||
\(\ds \) | \(\cdots\) | \(\ds \) | ||||||||||||
\(\ds n^3\) | \(=\) | \(\ds n + 2 n + 3 n + \cdots + \paren {n - 1} \paren n + n^2 + \paren {n - 1} \paren n + \cdots + 2 n + n\) | 1+2+...+n+(n-1)+...+1 = n^2 |
Therefore:
- $\forall n \in \Z_{>0}: \ds \sum_{i \mathop = 1}^n i^3 = \dfrac {n^2 \paren {n + 1}^2} 4$
$\blacksquare$
Visual Demonstration
A visual illustration of the proof for $n = 5$:
The number of squares of side $n$ is seen to be $4 n$.
To go from $n$ to $n + 1$, a ring of $4 \paren {n + 1}$ squares of side $n + 1$ is to be added:
- $4$ for the one at each corner
- $4 n$ for the ones that abut the sides of the $n + 1$ squares of side $n$.
The length of one side is given by:
- $S = 2 \paren {1 + 2 + \cdots + n}$
The length of one side is also given by:
- $S = n \paren {n + 1}$
The area is therefore given in two ways as:
- $A = 4 \paren {1 + 2 + \cdots + n}^2 = \paren {n \paren {n + 1} }^2$
and also as:
\(\ds A\) | \(=\) | \(\ds 4 \times 1^2 + 4 \times 2 \times 2^2 + 4 \times 3 \times 3^2 + \cdots + 4 \times n \times n^2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 4 \paren {1^3 + 2^2 + 3^3 + \cdots + n^3}\) |
The result follows by equating the expressions for area.
$\blacksquare$
Sequence
The sequence of integers which are the sum of the sequence of the first $n$ cubes begins:
- $0, 1, 9, 36, 100, 225, 441, 784, 1296, 2025, 3025, \ldots$
Also presented as
The Sum of Sequence of Cubes can also be presented as:
- $\ds \sum_{i \mathop = 0}^n i^3 = \paren {\sum_{i \mathop = 0}^n i}^2 = \frac {n^2 \paren {n + 1}^2} 4$
This is seen to be equivalent to the given form by the fact that the first term evaluates to $\dfrac {0^2 \paren {0 + 1}^2 } 4$ which is zero.
Examples
36
- $36 = 1^3 + 2^3 + 3^3 = 6^2 = \paren {1 + 2 + 3}^2$
100
- $100 = 1^3 + 2^3 + 3^3 + 4^3 = 10^2 = \left({1 + 2 + 3 + 4}\right)^2$
225
- $225 = 1^3 + 2^3 + 3^3 + 4^3 + 5^3 = 15^2 = \paren {1 + 2 + 3 + 4 + 5}^2$
Also see
Historical Note
The result Sum of Sequence of Cubes was documented by Aryabhata the Elder in his work Aryabhatiya of $499$ CE.
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 18$: Induced $N$-ary Operations: Exercise $18.10 \ \text{(c)}$
- 1971: George E. Andrews: Number Theory ... (previous) ... (next): $\text {1-1}$ Principle of Mathematical Induction: Exercise $2$
- 1971: Robert H. Kasriel: Undergraduate Topology ... (previous) ... (next): $\S 1.17$: Finite Induction and Well-Ordering for Positive Integers: Exercise $4$
- 1977: Gary Chartrand: Introductory Graph Theory ... (previous) ... (next): Appendix $\text{A}.6$: Mathematical Induction: Problem Set $\text{A}.6$: $37$
- 1982: P.M. Cohn: Algebra Volume 1 (2nd ed.) ... (previous) ... (next): Chapter $2$: Integers and natural numbers: $\S 2.1$: The integers: Exercise $11$