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### Calculating a Series

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Date: 1/23/96 at 16:23:26
From: Anonymous
Subject: Formula help?

Dr. Math:
I have a question about finding the formula that will calculate a
series of numbers not starting with 1 and not necessarily increasing
by 1. For  example the formula should be able to calculate
5+6+7+8+9+10+11 and 4+7+10+13 and 3+5+.....+47 ( all odd
numbers between 3 and 47). Can you help me? My teacher
mentioned that the formula was developed by a math genius centuries
ago. I can't come up with anything. Oh, it  also should be an
extension or change of n(n+1)/2.

Any help would be greatly appreciated!
CAMS28
```

```
Date: 5/31/96 at 15:41:31
From: Doctor Charles
Subject: Re: Formula help?

Well first here's an explanation of the original formula which is
easy to generalise:

What is 1+2+3+...+n?

Well it's the same as  (1+n)+(2+(n-1))+(3+(n-2))+...+ (n/2 + n-(n/2-1))
if n is even and:           (1+n)+(2+(n-1))+(3+(n-2))+...+ ((n+1)/2) if
n is odd.

But each pair of numbers adds up to n+1 and there are n/2 pairs of
numbers (if n is odd this means that there is 'half a pair' of numbers
in the last bracket. This is okay as it 'adds up' to (n+1)/2, so we can
count it as half of a pair of numbers adding up to n+1.

In either case we have n/2 pairs of numbers adding up to n+1, so the
total is n/2 * n+1 = n(n+1)/2.

General case:  a + (a+d) + (a+2d) + ... + (a+(n-1)d).
This is just   (a + (a+(n-1)d)) + ((a+d) + (a+(n-2)d) + ....
But now we have each outer bracket adding up to 2a+(n-1)d and there
are n/2 brackets. Can you write down the formula now?

Once you have done that you might like to think of it as the number of
terms (that is, n) multiplied by the average of the first and last terms.

If a is the first term and l is the last term and there are n terms, then
we have :

Sum = n*(a+l)/2.

Can you get this to agree with your other formula and the original
formula?

-Doctor Charles,  The Math Forum

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Associated Topics:
High School Sequences, Series

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