This preview shows page 1. Sign up to view the full content.
Unformatted text preview: Problem 25, section 3.2 Ugh. This problem. Well, lets see. We want to find a sum for
m k k=0 For simplicity, we'll let n = m  1, so that m = n + 1. Why is n significant? Well, if you right out the sum
m k = 0 + 1 + 1 + 1 + 2 + 2 + 2 + 2 + 2 + 3 + 3 + ... k=0 you'll notice that n is the second to last term that appears (before we start adding m . This is important because we don't know how many times we add m . But we do now how many times we add everything below it. As you m can see above, this fromula looks a lot like k=0 (2k + 1)k. But is isn't quite the same, because we may not add m as much as possible. Thus, we need to break this sum into two parts:
n (2k + 1)k +
i=0 m You'll notice the sum on the left is relatively straight forward, and if you work it out, it is equal to n(n+1)(2n+1) + n(n+1) . Thus we only need to figure 6 2 out the second part. You'll notice i left the second without any bounds for the sum. That's because we need to don't really need them: we are just adding m a bunch of times, so we just need to figure out how many times that is. Well, we just need to find the biggest square less than m. Then the difference between m and biggest square less than m is the number of times we add that m . Well, ( m )2 is a square that is less than m. Further, it should be obvious that this (which is equal to (n + 1)2 ) is the largest square less than m. m a total of m  (n + 1)2 + 1 times (we need the +1 Thus, we need to add because we add m one more time at the end, when we plug in m. Thus we get the answer in the book: n(n + 1)(2n + 1) n(n + 1) + + (n + 1)(m  (n + 1)2 + 1) 6 2 1 ...
View
Full
Document
This note was uploaded on 08/11/2008 for the course MATH 240 taught by Professor Miller during the Fall '08 term at Wisconsin.
 Fall '08
 Miller
 Math

Click to edit the document details