Closed Form for Triangular Numbers/Proof using Bernoulli Numbers

From ProofWiki
Jump to navigation Jump to search

Theorem

The closed-form expression for the $n$th triangular number is:

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


Proof

From Sum of Powers of Positive Integers:

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

where $B_k$ are the Bernoulli numbers.


Setting $p = 1$:

\(\displaystyle \sum_{i \mathop = 1}^n i^1\) \(=\) \(\displaystyle 1 + 2 + \cdots + n\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {n^2} 2 + \frac {B_1 \, p^{\underline 0} n^1} {1!}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {n^2} 2 + \frac 1 2 \frac {1 \times n} 1\) Definition of Bernoulli Numbers, Number to Power of Zero Falling is One
\(\displaystyle \) \(=\) \(\displaystyle \frac {n^2 + n} 2\)

Hence the result.

$\blacksquare$


Sources