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Topic: Equal sums of terms in Arithmetic Progession
Replies: 1   Last Post: Dec 28, 2012 6:56 PM

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 Philippe 92 Posts: 15 Registered: 4/16/12
Re: Equal sums of terms in Arithmetic Progession
Posted: Dec 28, 2012 6:56 PM

croomejohn@yahoo.co.uk a écrit :
> 1+2+...+14 = 15+16+...+20
>
> 1+4+7+10+13 = 16+19
>
> 1+6+11+...+251 = 256+261+...+356
>
> 1+7+13+...+55 = 61+67+73+79
>
> So with d = 1,3,5 and 6 (and 10 and 16, I haven't shown these), there exists
> n and m such that
>
> 1 + (1+d) + ... + 1+ nd = 1+(n+1)d + ... + 1+ md
>
> It is impossible with d = 2. Owing to 1+3+...+2n+1 = n^2 and 1+3+...2m+1 =
> m^2 so m^2 = 2n^2 leads to root(2) being rational m/n.

hi,

but the right member is not "1 +..." but "a+(a+d)+(a+2d)+...+(a+md)"

so that for instance :
1+3+5+7+...+15 = 64 = 100 - 36 = 13 + 15 + 17 + 19

with d = 2 !

Regards.

--
Philippe C. mail chephip à free.fr
site divertissement mathématiques http://mathafou.free.fr

Date Subject Author
12/28/12 croomejohn@yahoo.co.uk
12/28/12 Philippe 92