Summing Integers to the Fourth PowerDate: 09/26/98 at 13:20:49 From: Michael Subject: Formula for series of numbers to the power 4 I need to find a sum for the following series: 1^4 1^4 + 2^4 1^4 + 2^4 + 3^4 ... where the answers are 1, 17, 98, .... Michael Date: 09/26/98 at 16:28:04 From: Doctor Floor Subject: Re: Formula for series of numbers to the power 4 Hi Michael, Thank you for sending your question to Dr. Math! There is a formula for the series, which is given by: 1^4 + 2^4 + ... + n^4 = (1/5)n^5 + (1/2)n^4 + (1/3)n^3 - (1/30)n It seems magic where this comes from, but an explanation can be found in our archives. The explanation is written in such a way that you can also apply it to other questions like this one. See: http://mathforum.org/dr.math/problems/kyungsoo6.28.98.html If you have a math question again, send it to Dr. Math. Best regards, - Doctor Floor, The Math Forum http://mathforum.org/dr.math/ Date: 09/26/98 at 17:04:34 From: Doctor Mitteldorf Subject: Re: Formula for series of numbers to the power 4 Dear Michael, We should really start with the answers for the simpler problems: 1) What is the sum of the integers 1, 2, 3, 4, ..., n? To get this one, you add the first and the last, the second and the next to last, and so on. Let's try it for n = 8: 1 + 8 = 9 2 + 7 = 9 3 + 6 = 9 4 + 5 = 9 So we have 4 pairs, and each pair adds up to 9. In general, you can see we will have n/2 pairs, and each pair adds up to n+1. So the sum of the numbers from 1 to n is n*(n+1)/2. (I'm going to write powers as n^2 means n squared and n^4 means n to the fourth power.) 2) What is the sum of the squares of the integers: 1, 4, 9, 16, ..., n^2 This is a much harder problem, but here's one way to answer it: you can guess that the solution is an^3 + bn^2 + cn + d, but we still don't know what a,b,c and d are. To find a,b,c and d you can use this trick: assume you know that the sum of the squares from 1 to n is an^3 + bn^2 + cn + d. Then you can write the sum of the squares from 1 to (n+1) in two ways: first it's just an^3 + bn^2 + cn + d + (n+1)^2. Second, it's a(n+1)^3 + b(n+1)^2 + c^(n+1) + d. You can use this equation to find the four numbers, a,b,c, and d, then do some algebra to make your answer look like this: n(n+1)(2n+1)/6 3) For the cubes, it's the same idea but more algebra. The answer is (n^2 (n+1)^2)/4. This answer and the previous ones are in standard math table books. 4) For the sum of the fourth powers, it's not harder, just longer. The answer is n^5/5 + n^4/2 + n^3/3 - n/30. When you write it this way, of course, it isn't even obvious that the result is an integer, so it would be nice to have it in a factored form like the others, but I haven't taken the time to do that, and since this answer isn't in the standard books that I have I can't look it up. If you get a nice factored form, will you send it to me? - Doctor Mitteldorf, The Math Forum http://mathforum.org/dr.math/ |
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