Date: Wed, 9 Apr 1997 17:49:27 -0400

From: Alexander Bogomolny

The series consists of

one 1, two 2s, three 3s, ...

The sum of all numbers from 1 through n equals S_{n} = n(n+1)/2. You are looking for an n such that S_{n} < 1000000 but S_{n+1} > 1000000. This number is 1413.

How do you get it? n*(n+1) < (n+1)^{2} < 2000000.

sqrt(2000000) is approximately 1414.2... So n+1=1414.

Regards

64850408 |