Search All of the Math Forum:
Views expressed in these public forums are not endorsed by
NCTM or The Math Forum.



Equal sums of terms in Arithmetic Progession
Posted:
Dec 28, 2012 5:18 PM


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.



