Sum of Sequence of Triangular Numbers/Proof 2
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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
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 | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {2 m \paren {2 m + 1} \paren {2 m + 2} + \paren {6 m + 3} \paren {2 m + 2} } 6\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\paren {2 m + 2} \paren {2 m \paren {2 m + 1} + \paren {6 m + 3} } } 6\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\paren {2 m + 2} \paren {4 m^2 + 8 m + 3} } 6\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\paren {2 m + 2} \paren {2 m + 1} \paren {2 m + 3} } 6\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {n \paren {n + 1} \paren {n + 2} } 6\) |
$\blacksquare$
Sources
- 1980: David M. Burton: Elementary Number Theory (revised ed.) ... (previous) ... (next): Chapter $1$: Some Preliminary Considerations: $1.3$ Early Number Theory: Problems $1.3$: $3$