Sum of Sequence of Cubes/Proof using Bernoulli Numbers

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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

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.

$\blacksquare$


Also see


Sources

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