Search All of the Math Forum:
Views expressed in these public forums are not endorsed by
Drexel University or The Math Forum.
|
|
|
|
Re: Equal sums of terms in Arithmetic Progession
Posted:
Dec 28, 2012 7:34 PM
|
|
On Sat, 29 Dec 2012 00:56:24 +0100, Philippe 92 wrote:
> 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.
> 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 !
True, although that example doesn't start out on the right side
with the next term after the last one on the left side. The
algebra for m^2 = 2n^2 is that LHS = n^2, RHS = m^2 - n^2,
LHS = RHS, so m^2 = 2n^2.
--
jiw
|
|
|
|
