General Approach for Sum of Arithmetic SeriesDate: 08/24/2006 at 17:46:55 From: robert Subject: algebra/geometry What is the formula for figuring out the sum of adding consecutive even numbers? 2 + 4 = 6, 2 + 4 + 6 = 12, 2 + 4 + 6 + 8 = 20, etc. Date: 08/25/2006 at 14:54:39 From: Doctor Greenie Subject: Re: algebra/geometry Hi, Robert - You can memorize the formula for this particular type of problem; or you can learn a (fairly easy) general method for finding the sum of ANY sequence of numbers in which the difference between successive numbers is a constant amount. The quick formulas for special cases, including yours, are the following: (1) Sum of first n odd integers: n^2 1 = 1 = 1^2 1+3 = 4 = 2^2 1+3+5 = 9 = 3^2 1+3+5+7 = 16 = 4^2 ... (2) Sum of first n even integers: n^2 + n; or n(n + 1) 2 = 1(2) 2+4 = 6 = 2(3) 2+4+6 = 12 = 3(4) 2+4+6+8 = 20 = 4(5) ... (3) Sum of first n integers: (n(n + 1))/2 1 = 1(2)/2 1+2 = 3 = 2(3)/2 1+2+3 = 6 = 3(4)/2 1+2+3+4 = 10 = 4(5)/2 ... All of these are special cases of a general formula which I think is easy to remember: Sum = (number of numbers) * (first + last)/2 In the second part of this formula, we are adding two numbers and dividing by 2; that means we are taking the average of the two numbers. So I like to think of this formula as Sum = (number of numbers) * (average of first and last) In any sequence of numbers in which the difference between successive terms is the same, the average of all the numbers is the same as the average of the first and last numbers. So our formula really just says that the sum of a bunch of numbers is equal to the average, multiplied by how many there are. This makes perfect sense, since an average is found by dividing the total by how many there are. Let's look at a couple of the special case formulas and see that they are indeed particular examples of this general formula. (2 again) Sum of first n even integers 2+4+6+...+2n There are n numbers; the average of the first and last is (2n+2)/2 = n+1 And so our general formula gives us the special case formula: S = n(n+1) (3 again) Sum of first n integers 1+2+3+...+n There are again n numbers; the average of the first and last is (n+1)/2. The general formula again gives us the special case formula: S = n(n+1)/2 It certainly doesn't hurt to memorize the special case formulas. But you can use the general formula to find sums of vastly different sequences. For example... 90+95+100+105+...+235+240 The difference between the first and last terms is 240 - 90 = 150; the common difference between terms is 5; so there are 150/5 = 30 terms after the first one; so the number of terms in the sequence is 31. The average of the first and last terms is (90+240)/2 = 330/2 = 165. So the sum of this sequence of numbers is 31(165) = 5115 I hope this helps. Please write back if you have any further questions about any of this. - Doctor Greenie, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/