Closed Form for Triangular Numbers/Combinatorial Proof

Theorem

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

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

Combinatorial Proof

This article needs to be tidied. In particular: Please review our site presentational standards and protocols. Please fix formatting and $\LaTeX$ errors and inconsistencies. It may also need to be brought up to our standard house style. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Tidy}} from the code.

Suppose we have $n + 1$ people who all shake hands with each other person once then we can count how many handshakes occur by counting the handshakes of each person, the first person shakes hands with all other $n$ people, then the second person shakes hands with the remaining $n - 1$ people, which continues until the $n$-th person can only shake hands with the $(n + 1)$-th person, so counting backwards this consists of $\sum_{i = 0}^n i$ handshakes. Similarly, we can count the number of choices of handshakes between any two of the $n + 1$ people, which will be ${n + 1 \choose 2} = \frac{n(n + 1)}{2}$ handshakes, which is our proposed solution to the sum.