Calculating a SeriesDate: 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 |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/