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Give an iterative algorithm and loop invariant for 1+2+3+4+........+n
it must be iterative(loop)
3 Answers
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print(sum(range(n+1)))
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#if iterative implies recursive
s=lambda x: (0 if x==0 else x+s(x-int(abs(x)/x)))
#recursive
print(s(10))
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a loop version of this algorithm won't do you well on arbitrarily big numbers (billions and up) so if the need for it arises, a pure math solution is avalible:
(1+2+3+...+n) == (n*(n+1))/2