Closed Form for Triangular Numbers/Direct Proof

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

We have that:

$\displaystyle \sum_{i \mathop = 1}^n i = 1 + 2 + \cdots + n$

Consider $\displaystyle 2 \sum_{i \mathop = 1}^n i$.

Then:

\(\displaystyle 2 \sum_{i \mathop = 1}^n i\) \(=\) \(\displaystyle 2 \paren {1 + 2 + \dotsb + \paren {n - 1} + n}\)
\(\displaystyle \) \(=\) \(\displaystyle \paren {1 + 2 + \dotsb + \paren {n - 1} + n} + \paren {n + \paren {n - 1} + \dotsb + 2 + 1}\)
\(\displaystyle \) \(=\) \(\displaystyle \paren {1 + n} + \paren {2 + \paren {n - 1} } + \dotsb + \paren {\paren {n - 1} + 2} + \paren {n + 1}\) Integer Addition is Commutative, Integer Addition is Associative
\(\displaystyle \) \(=\) \(\displaystyle \paren {n + 1}_1 + \paren {n + 1}_2 + \dotsb + \paren {n + 1}_n\)
\(\displaystyle \) \(=\) \(\displaystyle n \paren {n + 1}\)

So:

\(\displaystyle 2 \sum_{i \mathop = 1}^n i\) \(=\) \(\displaystyle n \paren {n + 1}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \sum_{i \mathop = 1}^n i\) \(=\) \(\displaystyle \frac {n \paren {n + 1} } 2\)

$\blacksquare$


Historical Note

This direct proof is the method supposedly employed by Carl Friedrich Gauss who, when very young (according to the apocryphal story), calculated the sum of the numbers from $1$ to $100$ before the teacher had barely sat back down after setting the assignment.

Whether this story is actually true or not is the subject of speculation.


Sources