((n+1)/2)nDate: 01/31/2002 at 05:52:53 From: Alan Naude Subject: Proof of an equation If you want to figure out the total of a series of numbers in order, e.g.: 1+2+3+4+5+6+7+8+9 etc., you would use the formula below where n is the final number of your series. ((n+1)/2)n Why, and what is the proof of the formula? Date: 01/31/2002 at 08:28:51 From: Doctor Jerry Subject: Re: Proof of an equation Hi Alan, Any finite sum of the form S = a + (a+d) + (a+2d) + (a+3d) + ... + (a+(n-1)d) can be summed. For the series you mentioned, take a = 1 and d = 1. Write the sum as above and reverse it: S = a + (a+d) + (a+2d) + (a+3d) + ... + (a+(n-1)d) S = (a+(n-1)d) + (a+(n-2)d) + ...+(a+d) + a Add together: 2S = n*[2a+(n-1)d] So, S = n[a + a + (n-1)d]/2. So, to sum a finite arithmetic progression, average the first and last terms and multiply by the number n of terms. - Doctor Jerry, The Math Forum http://mathforum.org/dr.math/ |
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