Determine all possible sequences of consecutive triangular numbers whose sum is precisely 2014.
Extra Challenge: A non computer program based method.
Consider a 'sequence of consecutive triangular numbers' as the difference of two sequences, both starting at 1, and ending at m and n respectively.
Then 1/2n(n+1)1/2m(m+1) = 1/8 (2n+1)^21/8(2m+1)^2 , with 0<m<n, describes all possible such truncated sequences.
So (2n+1)^2(2m+1)^2 = 8*2014 = 16112
(2n2m)(2n+2m+2) = 16112
(nm) (m+n+1) = 4028 = 2^2*19*53
It's then just a matter of plugging in the factors:
nm=1, n+m+1=4028; m=2013,n=2014
nm=4, n+m+1=1007; m=501, n=505
nm=19, n+m+1=848; m=96 n=115
nm=53, n+m+1= 76; m=11, n=64
There are no other positive integer solutions.
Edited on August 12, 2014, 3:31 pm

Posted by broll
on 20140812 15:26:33 