Closed Form for Triangular Numbers/Proof using Bernoulli Numbers

Theorem

$\ds T_n = \sum_{i \mathop = 1}^n i = \frac {n \paren {n + 1} } 2$

$\ds \sum_{i \mathop = 1}^n i^p$

$=$

$\ds 1^p + 2^p + \cdots + n^p$

$\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 $p = 1$:

$\ds \sum_{i \mathop = 1}^n i^1$

$=$

$\ds 1 + 2 + \cdots + n$

$\ds $

$=$

$\ds \frac {n^2} 2 + \frac {B_1 \, p^{\underline 0} n^1} {1!}$

$\ds $

$=$

$\ds \frac {n^2} 2 + \frac 1 2 \frac {1 \times n} 1$

$\ds $

$=$

$\ds \frac {n^2 + n} 2$

Hence the result.

$\blacksquare$