# Sum of Sequence of Triangular Numbers

## Theorem

Let $n \in \Z_{>0}$ be a (strictly) positive integer.

Let $T_n$ denote the $n$th triangular number.

Then:

 $\ds \sum_{j \mathop = 1}^n T_j$ $=$ $\ds T_1 + T_2 + T_3 + \dotsb + T_n$ $\ds$ $=$ $\ds \dfrac {n \paren {n + 1} \paren {n + 2} } 6$

## Proof 1

From Sum of Sequence of n Choose 2 we have:

 $\ds \sum_{j \mathop = 2}^n \dbinom j 2$ $=$ $\ds \dbinom 2 2 + \dbinom 3 2 + \dbinom 4 2 + \dotsb + \dbinom n 2$ $\ds$ $=$ $\ds \dbinom {n + 1} 3$

and so:

 $\ds \sum_{j \mathop = 2}^{n + 1} \dbinom j 2$ $=$ $\ds \dbinom 2 2 + \dbinom 3 2 + \dbinom 4 2 + \dotsb + \dbinom n 2 + \dbinom {n + 1} 2$ $\ds$ $=$ $\ds \dbinom {n + 2} 3$

But we have that:

$\dbinom {n + 2} 3 = \dfrac {\paren {n + 2} \paren {n + 1} n} {3 \times 2 \times 1}$

and from Binomial Coefficient with Two:

$T_n = \dfrac {\paren {n + 1} n} 2 = \dbinom {n + 1} 2$

The result follows.

$\blacksquare$

## Proof 2

First let $n$ be even.

Thus we have:

$n = 2 m$

Then:

 $\ds T_1 + T_2 + T_3 + \dotsb + T_{2 m}$ $=$ $\ds \paren {T_1 + T_2} + \paren {T_3 + T_4} + \dotsb + \paren {T_{2 m - 1} + T_{2 m} }$ $\ds$ $=$ $\ds 2^2 + 4^2 + \dotsb + \paren {2 m}^2$ Sum of Consecutive Triangular Numbers is Square $\ds$ $=$ $\ds 2^2 \times 1^2 + 2^2 \times 2^2 + \dotsb + 2^2 \times m^2$ Sum of Consecutive Triangular Numbers is Square $\ds$ $=$ $\ds 2^2 \paren {1^2 + 2^2 + \dotsb + m^2}$ $\ds$ $=$ $\ds 4 \frac {m \paren {m + 1} \paren {2 m + 1} } 6$ $\ds$ $=$ $\ds \frac {2 m \paren {2 m + 1} \paren {2 m + 2} } 6$ $\ds$ $=$ $\ds \frac {n \paren {n + 1} \paren {n + 2} } 6$

Now let $n$ be odd.

Thus we have:

$n = 2 m + 1$

Then:

 $\ds T_1 + T_2 + T_3 + \dotsb + T_{2 m}$ $=$ $\ds \paren {T_1 + T_2} + \paren {T_3 + T_4} + \dotsb + \paren {T_{2 m - 1} + T_{2 m} } + T_{2 m + 1}$ $\ds$ $=$ $\ds \frac {2 m \paren {2 m + 1} \paren {2 m + 2} } 6 + T_{2 m + 1}$ from above $\ds$ $=$ $\ds \frac {2 m \paren {2 m + 1} \paren {2 m + 2} } 6 + \frac {\paren {2 m + 1} \paren {2 m + 2} } 2$ Closed Form for Triangular Numbers $\ds$ $=$ $\ds \frac {2 m \paren {2 m + 1} \paren {2 m + 2} + \paren {6 m + 3} \paren {2 m + 2} } 6$ $\ds$ $=$ $\ds \frac {\paren {2 m + 2} \paren {2 m \paren {2 m + 1} + \paren {6 m + 3} } } 6$ $\ds$ $=$ $\ds \frac {\paren {2 m + 2} \paren {4 m^2 + 8 m + 3} } 6$ $\ds$ $=$ $\ds \frac {\paren {2 m + 2} \paren {2 m + 1} \paren {2 m + 3} } 6$ $\ds$ $=$ $\ds \frac {n \paren {n + 1} \paren {n + 2} } 6$

$\blacksquare$

## Historical Note

David M. Burton, in his Elementary Number Theory, revised ed. of $1980$, reports that this result is attributed to Aryabhata the Elder and dates from circa $500$ C.E.