Determine all possible sequences of consecutive triangular numbers whose sum is precisely 2014.

Extra Challenge: A non computer program based method.

(In reply to re: Possible Solution by Charlie)

Because I answered the wrong question, Charlie!

(n-m) (m^2+mn+3m+n^2+3n+2) = 12084 = 2^2×3×19×53 gives:

1. ((n-m)=19:) m=3, n=22 (i.e. T(4)...T(22))

2. ((n-m)=2^2*3:) m=11, n=23 (i.e. T(12)...T(23))

corresponding with your solution.

Edited on August 13, 2014, 2:07 am

Posted by broll on 2014-08-13 01:13:26