Sum of Sequence of Cubes

From ProofWiki
Jump to navigation Jump to search

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 Sum of Powers of Positive Integers:

\(\ds \sum_{i \mathop = 1}^n i^p\) \(=\) \(\ds 1^p + 2^p + \cdots + n^p\)
\(\ds \) \(=\) \(\ds \frac {n^{p + 1} } {p + 1} + \sum_{k \mathop = 1}^p \frac {B_k \, p^{\underline {k - 1} } \, n^{p - k + 1} } {k!}\)

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 {\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 $n = 5$:


SumOfCubes-Floyd.png


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

Retrieved from "https://proofwiki.org/w/index.php?title=Sum_of_Sequence_of_Cubes&oldid=614088"