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 lgchic April 2nd, 2008 07:15 PM

Sum of arithmetic series...

Consider a finite arithmetic series which has the form:

x + (x+d) + (x+2d) + (x+3d) + ... + (x+(n-1)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 1-1000. 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!

 CRGreathouse April 3rd, 2008 05:11 AM

Re: Sum of arithmetic series...

Quote:
 Originally Posted by lgchic Consider a finite arithmetic series which has the form: x + (x+d) + (x+2d) + (x+3d) + ... + (x+(n-1)d) where x is the initial value and d is the common difference. Find the sum of this series.
There are n x's, so you can pull those out and replace them with nx:
nx + [0 + d + 2d + ... + (n-1)d]

You can then factor out d:
nx + d[0 + 1 + 2 + ... + (n-1)]

Since you know how to sum 1 + 2 + ... + n, I imagine you can solve this now?

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