April 2nd, 2008, 07:15 PM  #1 
Newbie Joined: Apr 2008 Posts: 1 Thanks: 0  Sum of arithmetic series...
Consider a finite arithmetic series which has the form: x + (x+d) + (x+2d) + (x+3d) + ... + (x+(n1)d) where x is the initial value and d is the common difference. Find the sum of this series. We're given the hint: suppose you wanted to add up all the numbers from 11000. The standard trick is to write the series first in ascending order, then in descending order and then add the corresponding terms like so: First ascending: 1, 2, 3, ... 999, 1000 Then descending: 1000, 999, 998 ... 2, 1 Now add: 1+1000=1001, 2+999=1001, 3+998=1001, ... 999+2=1001, 1000+1=1001 Now notice that the sum of all the terms in the last line is easy to calculate and, it's exactly twice the quantity we seek. So: 1+2+3+ ... +999+1000= [1000(1001)/2] = 500500 Use this same method to answer the more general question. Any help would be greatly appreciated! Thank you! 
April 3rd, 2008, 05:11 AM  #2  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Re: Sum of arithmetic series... Quote:
nx + [0 + d + 2d + ... + (n1)d] You can then factor out d: nx + d[0 + 1 + 2 + ... + (n1)] Since you know how to sum 1 + 2 + ... + n, I imagine you can solve this now?  

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