Math 2.0

Combinatorics Theorem

Sum of the first n natural numbers

Builds on Mathematical induction

Statement

For every natural number nn, 0+1+2+⋯+n=∑k=0nk=n(n+1)2.0 + 1 + 2 + \dots + n = \sum_{k=0}^{n} k = \frac{n(n + 1)}{2}.

Remarks

The number n(n+1)/2n(n + 1)/2 is the nnth triangular number. For n=1,2,3,4n = 1, 2, 3, 4 it is 1,3,6,101, 3, 6, 10. A proof by induction uses n(n+1)2+(n+1)=(n+1)(n+2)2\frac{n(n+1)}{2} + (n + 1) = \frac{(n+1)(n+2)}{2}. An equivalent form without division is 2∑k=0nk=n(n+1)2 \sum_{k=0}^{n} k = n(n + 1).

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Where this leads

2 pages build on this one.

Further reading

  1. Web https://en.wikipedia.org/wiki/Triangular_number, Formula
  2. Web https://en.wikipedia.org/wiki/Mathematical_induction, Sum of consecutive natural numbers
About this page
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Kindstatement, theorem
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MSC05A19
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UsesMathematical induction

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