Difference between revisions of "Triangular number"

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Using the sum of an [[arithmetic series]] formula, a formula can be calculated for <math>\displaystyle T_n</math>:
 
Using the sum of an [[arithmetic series]] formula, a formula can be calculated for <math>\displaystyle T_n</math>:
  
:<math>T_n = \displaystyle\sum_{i=1}n = 1 + 2 \cdots n = \frac{n(n+1)}2</math>
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:<math>T_n = \displaystyle\sum_{k=1}^{n}k = 1 + 2 + \ldots + n = \frac{n(n+1)}2</math>
  
The rather simple recursive definition can be easily found by noting that <math>\displaystyle T_{n} = 1 + 2 \cdots (n-1) + n = (1 + 2 \cdots n-1) + n = T_{n-1} + n</math>.
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The rather simple recursive definition can be easily found by noting that <math>\displaystyle T_{n} = 1 + 2 + \ldots + (n-1) + n = (1 + 2 + \ldots + n-1) + n = T_{n-1} + n</math>.
  
 
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Revision as of 03:11, 8 September 2007

The triangular numbers are the numbers $\displaystyle T_n$ which are the sum of the first $\displaystyle n$ natural numbers from $\displaystyle 1$ to $\displaystyle n$.

Using the sum of an arithmetic series formula, a formula can be calculated for $\displaystyle T_n$:

$T_n = \displaystyle\sum_{k=1}^{n}k = 1 + 2 + \ldots + n = \frac{n(n+1)}2$

The rather simple recursive definition can be easily found by noting that $\displaystyle T_{n} = 1 + 2 + \ldots + (n-1) + n = (1 + 2 + \ldots + n-1) + n = T_{n-1} + n$.

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