Closed Form for Triangular Numbers

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

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

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

Consider $$2 \sum_{i=1}^{n} i$$. 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:
 * $$(i+1)^{2}-i^{2}=2i+1$$ and
 * $$(n+1)^{2}-1=n^{2}+2n$$

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

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

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

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

So the base case is true.

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

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

By the properties of summation,$$\sum_{i=1}^{k+1}i=\sum_{i=1}^{k}i +k+1$$.

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 $$T_n = \frac {n \left({2 + \left({n-1}\right)\left({k-2}\right)}\right)} 2$$ where $$k = 3$$.

The result follows after some simple algebra.