# Sum of Sequence of Triangular Numbers/Proof 1

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## 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

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.

$\blacksquare$