Thursday, May 9, 2013

P5 Sum of series (Fr CCL)

Hope that no one solves this question (b) in the way shown above! 

My solution:
 2011/8 = 251R3 

(a) Since the remainder is 3, the last number in this series is 3

The number of terms in this series is 251+1 = 252  ( The "+1" is to cater to the last tern "3")
(b)   The sum of the series 
= 252/2  x (2011 + 3)
=  126 x 2014
=    253 764