# Sum of Sequence of Cubes/Proof using Bernoulli Numbers

## 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

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

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 $\ds$ $=$ $\ds \frac {n^4} 4 + \frac {n^3} 2 + \frac {n^2} 4$ simplifying $\ds$ $=$ $\ds \frac {n^2 \left({n + 1}\right)^2} 4$ after algebra

Hence the result.

$\blacksquare$