Closed Form for Triangular Numbers

Theorem
The closed-form expression for the $n$th triangular number is:
 * $\displaystyle T_n = \sum_{i=1}^{n} i = \frac {n \left({n+1}\right)} {2}$

Plainly stated: the sum of the first $n$ natural numbers is equal to $\displaystyle \frac {n \left({n+1}\right)} 2$.

This formula pops up frequently in fields as differing as calculus and computer science, and it is elegant in its simplicity.

Direct Proof
We have that $\displaystyle \sum_{i=1}^n i = 1 + 2 + \cdots + n$.

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

So:

This is the method employed by 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.

Direct Proof by using telescoping sum
Observe that

Moreover:
 * $ \left({i + 1}\right)^2 - i^2 = 2 i + 1$

and
 * $ \left({n + 1}\right)^2 - 1 = n^2 + 2n$

Thus $\displaystyle 2 \sum_{i=1}^n i + n = n^2 + 2 n \implies 2 \sum_{i=1}^n i = n \left({n + 1}\right) \implies \sum_{i=1}^n i = \frac {n \left({n + 1}\right)} 2$ as required.

Direct Proof by using Recursion
Let $S \left({n}\right) = 1 + 2 + \cdots + n = \sum_{i=1}^n i$.

Proof by Induction
Base Case: $n=1$:

When $n=1$, we have $\displaystyle \sum_{i=1}^{1}i=1$.

Also, $\displaystyle \frac{n(n+1)}{2}=\frac{1(2)}{2}=1$.

So the base case is true.

Inductive Hypothesis::
 * $\displaystyle \sum_{i=1}^{k}i=\frac{k(k+1)} 2$ for $k>1$

Inductive Step: Consider $n=k+1$.

By the properties of summation:
 * $\displaystyle \sum_{i=1}^{k+1}i=\sum_{i=1}^{k}i +k+1$

Using the induction hypothesis this can be simplified to:

Thus, the result has been shown by induction.

This is usually the first proof by induction that a student mathematician experiences.

Proof by Arithmetic Progression
Follows directly from Sum of Arithmetic Progression putting $a = 1$ and $d = 1$.

Proof by Polygonal Numbers
Triangular numbers are $k$-gonal numbers where $k = 3$.

Hence from Closed Form for Polygonal Numbers we have that $\displaystyle T_n = \frac {n \left({2 + \left({n-1}\right)\left({k-2}\right)}\right)} 2$ where $k = 3$.

The result follows after some simple algebra.

Proof using Binomial Coefficients
From Properties of Binomial Coefficients: Particular Values, we have that:
 * $\displaystyle \forall k \in \Z, k > 0: \binom k 1 = k$

Thus: