Sum of Sequence of Cubes

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Theorem

$\displaystyle \sum_{i \mathop = 1}^n i^3 = \left({\sum_{i \mathop = 1}^n i}\right)^2 = \frac{n^2 \left({n + 1}\right)^2} 4$


Proof by Induction

First, from Closed Form for Triangular Numbers:

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

So:

$\displaystyle \left({\sum_{i \mathop = 1}^n i}\right)^2 = \frac{n^2 \left({n + 1}\right)^2} 4$


Next we use induction on $n$ to show that $\displaystyle \sum_{i \mathop = 1}^n i^3 = \frac{n^2 \left({n + 1}\right)^2} 4$.

The base case holds since $\displaystyle 1^3 = \frac{1 \left({1 + 1}\right)^2} 4$.

Now we need to show that if it holds for $n$, then it holds for $n + 1$.

\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \sum_{i \mathop = 1}^{n+1} i^3\) \(=\) \(\displaystyle \) \(\displaystyle \sum_{i \mathop = 1}^n i^3 + \left({n + 1}\right)^3\) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \) \(\displaystyle \frac{n^2 \left({n + 1}\right)^2} 4 + \left({n + 1}\right)^3\) \(\displaystyle \) \(\displaystyle \)          (by the induction hypothesis)          
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \) \(\displaystyle \frac{n^4 + 2 n^3 + n^2} 4 + \frac {4 n^3 + 12 n^2 + 12 n + 4} 4\) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \) \(\displaystyle \frac{n^4 + 6 n^3 + 13 n^2 + 12 n + 4} 4\) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \) \(\displaystyle \frac{\left({n + 1}\right)^2 \left({n + 2}\right)^2} 4\) \(\displaystyle \) \(\displaystyle \)                    

By the Principle of Mathematical Induction, the proof is complete.

$\blacksquare$


Proof by Nicomachus

By Nicomachus's Theorem, we have:

$\forall n \in \N_{>0}: n^3 = \left({n^2 - n + 1}\right) + \left({n^2 - n + 3}\right) + \ldots + \left({n^2 + n - 1}\right)$


Also by Nicomachus's Theorem, we have that the first term for $\left({n + 1}\right)^3$ is $2$ greater than the last term for $n^3$.


So if we add them all up together, we get:

\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \sum_{i \mathop = 1}^n i^3\) \(=\) \(\displaystyle \) \(\displaystyle \sum_{\stackrel {1 \mathop \le i \mathop \le n^2 \mathop + n \mathop - 1} {i \text { odd} } } i\) \(\displaystyle \) \(\displaystyle \)          ... the sum of all the odd numbers up to $n^2 + n - 1$          
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \) \(\displaystyle \sum_{i \mathop = 1}^{\frac {n^2 \mathop + n} 2} 2 i - 1\) \(\displaystyle \) \(\displaystyle \)          ... that is, the first $\dfrac {n^2 + n} 2$ odd numbers          
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \) \(\displaystyle \left({\frac {n^2 + n} 2}\right)^2\) \(\displaystyle \) \(\displaystyle \)          by the Odd Number Theorem          


Hence the result.

$\blacksquare$


Proof by Recursion

From Closed Form for Triangular Numbers‎:

$(1): \quad \displaystyle A \left({n}\right) := \sum_{i \mathop = 1}^n i = \frac{n \left({n + 1}\right)} 2$

From Sum of Sequence of Squares:

$(2): \quad \displaystyle B \left({n}\right) := \sum_{i \mathop = 1}^n i^2 = \frac{n \left({n + 1}\right) \left({2 n + 1}\right)} 6$


Let $\displaystyle S \left({n}\right) = \sum_{i \mathop = 1}^n i^3$.

Then:

\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle S \left({n}\right)\) \(=\) \(\displaystyle \) \(\displaystyle n^3 + \left({n - 1}\right)^3 + \left({n - 2}\right)^3 + \cdots + 2^3 + 1^3\) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \) \(\displaystyle \sum_{k \mathop = 0}^{n \mathop - 1} \left({n - k}\right)^3\) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \) \(\displaystyle \sum_{k \mathop = 0}^{n \mathop - 1} \left({n^3 - 3 n^2 k + 3 n k^2 - k^3}\right)\) \(\displaystyle \) \(\displaystyle \)                    
\(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \) \(\displaystyle n^4 - 3 n^2 \cdot A \left({n - 1}\right) + 3 n \cdot B \left({n - 1}\right) - S \left({n - 1}\right)\) \(\displaystyle \) \(\displaystyle \)                    
\((3):\)      \(\displaystyle \) \(\displaystyle \implies\) \(\displaystyle \) \(\displaystyle S \left({n}\right)\) \(=\) \(\displaystyle \) \(\displaystyle n^4 - 3n^2 \cdot \frac{n \left({n - 1}\right)} 2 + 3n \cdot \frac{n \left({n - 1}\right) \left({2 n - 1}\right)} 6 - S \left({n - 1}\right)\) \(\displaystyle \) \(\displaystyle \)          substituting from $(1)$ and $(2)$          
\((4):\)      \(\displaystyle \) \(\displaystyle \) \(\displaystyle \) \(\displaystyle S \left({n}\right)\) \(=\) \(\displaystyle \) \(\displaystyle n^3 + S \left({n - 1}\right)\) \(\displaystyle \) \(\displaystyle \)          recursive definition          
\(\displaystyle \) \(\displaystyle \implies\) \(\displaystyle \) \(\displaystyle 2 S \left({n}\right)\) \(=\) \(\displaystyle \) \(\displaystyle n^4 + n^3 - 3n^2 \cdot \frac{n \left({n - 1}\right)} 2 + 3n \cdot \frac{n \left({n - 1}\right) \left({2 n - 1}\right)} 6\) \(\displaystyle \) \(\displaystyle \)          adding $(3)$ and $(4)$          
\(\displaystyle \) \(\displaystyle \implies\) \(\displaystyle \) \(\displaystyle \) \(=\) \(\displaystyle \) \(\displaystyle \frac{n^2 \left({n + 1}\right)^2} 2\) \(\displaystyle \) \(\displaystyle \)          simplification          
\(\displaystyle \) \(\displaystyle \implies\) \(\displaystyle \) \(\displaystyle S \left({n}\right)\) \(=\) \(\displaystyle \) \(\displaystyle \frac{n^2 \left({n + 1}\right)^2} 4\) \(\displaystyle \) \(\displaystyle \)                    

$\blacksquare$


Historical Note

This result was documented by Āryabhaṭa in his work Āryabhaṭīya of 499 CE.


Sources