# A proof for the closed form of triangular numbers

31 Dec 2022

The triangular numbers

$T_n = \sum_{i=1}^n i, \text{for }n>0$

have the generally known closed form

$T_n = \frac{1}{2}n(n+1)$

This can be easily proven by induction or young Gauss’s regrouping of terms.

Calculating some sums, I found another way using the sum of the squares. Even though I haven’t found it on the web yet, it is probably well-known.

## Proof

We look at the sum of squares

$S = \sum_{i=0}^{n+1} i^2$

On the one hand, we can remove the last term from the series:

$S = (n+1)^2 + \sum_{i=0}^n i^2$

On the other hand, we can remove the first term and shift indices:

\begin{aligned} S &= 0 + \sum_{i=1}^{n+1} i^2 = \sum_{i=0}^n (i+1)^2 \\ &= \sum_{i=0}^n (i^2+2i+1) = \sum_{i=0}^n i^2+\sum_{i=0}^n 2i+\sum_{i=0}^n 1 \\ &= \sum_{i=0}^n i^2 + 2\sum_{i=0}^n i + (n+1) \end{aligned}

Equating both sides we get

\begin{aligned} &(n+1)^2 + \sum_{i=0}^n i^2 = \sum_{i=0}^n i^2 + 2\sum_{i=0}^n i + (n+1) \\ \Leftrightarrow &(n+1)^2 = n+1+ 2\sum_{i=0}^n i \\ \Leftrightarrow &2\sum_{i=0}^n i = (n+1)^2-(n+1) = n(n+1) \\ \Leftrightarrow &\sum_{i=0}^n i = \sum_{i=1}^n i = \frac{1}{2}n(n+1) \end{aligned}

QED

## Some thoughts

This is kind of similar to the proof by telescoping series. A similar idea to remove the first and last element of a series is often used to calculate the partial sums of the geometric series.

This proof does not give as much insight as Gauss’s proof, but it is relatively easy for a proof by manipulation of sums. It also generalizes to the sum of $i^k$ for $k \in \mathbb{N}$ by always using the sum of $i^{k+1}$ and the binomic expansion of $(i+1)^{k+1}$.