Sum of Sequence of Triangular Numbers
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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 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.