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We prove by induction on `n`

forallninN: (sum_{1 <= i <= n}i) = (n+1)n/2

The induction base holds because

(sum_{1 <= i <= 0}i) = 0 = (0+1)*0/2.

We take arbitrary `n` in **N** and assume

(1) (sum_{1 <= i <= n}i) = (n+1)n/2.

We have to show

(2) (sum_{1 <= i <= n+1}i) = ((n+1)+1)(n+1)/2.

Author: Wolfgang Schreiner

Last Modification: November 24, 1999