Sum of all the digits of the integers from 1 to 2008
Answers
Answered by
5
sum of n natural numbers =n(n+1)/2
therefore
sum of all digits from 1 to 2008 is
= 2008( 2008+1)/2
= 4034072/2
= 2017036
therefore
sum of all digits from 1 to 2008 is
= 2008( 2008+1)/2
= 4034072/2
= 2017036
Similar questions