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January 30, 2015

January 30, 2015

Posted by **HELP!!** on Wednesday, May 8, 2013 at 5:30am.

- Maths -
**Steve**, Wednesday, May 8, 2013 at 11:55amIf there are k summands, starting at n, we want n as small as possible.

n+(n+1)+...+(n+k) = 1000

kn + k(k-1)/2 = 1000

k^2 + (2n-1)k - 2000 = 0

so

4n^2-4n+8001 is a perfect square

If n=28, 4n^2+4n+8001=105^2 and k=25

check:

summing the arithmetic sequence starting at 28 for 25 terms,

S25 = 25/2 (28*2+24) = 25/2*80 = 1000

So, it looks like 25 is the maximum number of summands.

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