Closed Form for Triangular Numbers/Proof using Bernoulli Numbers
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Theorem
The closed-form expression for the $n$th triangular number is:
- $\ds T_n = \sum_{i \mathop = 1}^n i = \frac {n \paren {n + 1} } 2$
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 $p = 1$:
\(\ds \sum_{i \mathop = 1}^n i^1\) | \(=\) | \(\ds 1 + 2 + \cdots + n\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {n^2} 2 + \frac {B_1 \, p^{\underline 0} n^1} {1!}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {n^2} 2 + \frac 1 2 \frac {1 \times n} 1\) | Definition of Bernoulli Numbers, Number to Power of Zero Falling is One | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {n^2 + n} 2\) |
Hence the result.
$\blacksquare$
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 19$: Sums of Powers of Positive Integers: $19.9$