
#1
Apr2510, 10:22 PM

P: 276

Hello,
I'm having some issues solving some apparently 'basic' summation problems where they give you a couple summations and you derive the missing summation. I would appreciate any help not only solving this particular question but actually understanding the situation. Thanks! 1. The problem statement, all variables and given/known data Suppose: 113 [tex]\Sigma[/tex] a_{k} = 641 k=9 and 507 [tex]\Sigma[/tex] a_{k} = 2091 k=70 and 113 [tex]\Sigma[/tex] a_{k} = 130 k=70 Then: 507 [tex]\Sigma[/tex] a_{k} = ? k=114 2. Relevant equations I think everything relevant is provided above. 3. The attempt at a solution I have absolutely no idea how to start. I've googled summation problems and read quite a bit more than what is in my text. I keep finding information on what I can already do; problems like: 10 [tex]\Sigma[/tex] a20 = ? k=1 Where a is a constant instead of having the 'k' subscript so you can take (10(11)) * (a20). Although, for some reason sticking the subscript onto a throws me off completely. Any suggestions are welcome. Thank you for your time! 



#2
Apr2510, 10:38 PM

HW Helper
P: 6,210

Draw it out.
70 113 114 507 So you have the sums from 70 to 113 and 70 to 507. How do you think you will get the required sum? 



#3
Apr2510, 10:56 PM

P: 276

Indeed you've helped me find the solution! Thank you
I take the summation from 70507 (2091) and subtract the summation from 70113 (130). Although, I guess I'm still not quite sure what's going on with the a_{k} portion. I'd like to be able to better understand how to evaluate each particular summation listed above rather than taking the short route to the answer in this particular situation. It may not always be so clear. Is there a way to evaluate those summations? Thanks again! 



#4
Apr2510, 11:04 PM

HW Helper
P: 2,156

Summations
I actually don't think so.
The subscript k is used to indicate different terms in the series. So for example, in the series of Fibbonacci numbers, you would have a_{1} = 1 a_{2} = 1 a_{3} = 2 a_{4} = 3 a_{5} = 5 and so on. a_{k} stands for a generic term, the kth term in the series. Here a_{k} would be the kth Fibbonacci number. Or in the series of prime numbers, you'd have a_{1} = 2 a_{2} = 3 a_{3} = 5 a_{4} = 7 a_{5} = 11 and so on. Here a_{k} would be the kth prime number. The terms of a series can be decreasing, can be nonintegers, negative, irrational, or whatever... for example a_{1} = 1 a_{2} = 1/2 a_{3} = 1/4 a_{4} = 1/8 a_{5} = 1/16 etc. In this case, you could write a formula for the kth term in the series, [tex]a_k = \frac{1}{2^{k1}}[/tex] (check it!) but in general that's not true. A series could even consist of completely random numbers. Generally you know absolutely nothing about the value of a_{k} for any k, unless someone tells you what the numbers in the series are. And unless that's the case, you can't find the sum of the series. 



#5
Apr2510, 11:07 PM

P: 276

Okay, so using the values I'm given for each of those summations I wouldn't be able to find 'a' without an unreasonable amount of work?
EDIT: Also, for something like 58 [tex]\Sigma[/tex] (35k19) = ? k=4 I understand that this is the sum of (35(4)19) + (35(5)19) + ... + (35(58)19) However, I can't find a way to derive a reasonable equation to calculate this without doing it all manually. I'm supposed to be able to do these on the test with a standard calculator without the sequence/summation functions built in. I doubt the goal here is for me to do 55 calculations and add them all. I can find the answer without a problem using a graphing calculator. Although, I would like to derive an equation that would let me manage the calculation on a simple calculator. Is that possible? 



#6
Apr2510, 11:45 PM

P: 351

You can't find the values of a_{k} at all. They're not guaranteed to fall into a nice pattern. The point is that you know the sums from various numbers to various other numbers, and so by judiciously adding and subtracting all the sums you can get certain other sums even though they aren't given.
If you need to evaluate something a sum like that, you can use the fact that [tex]\sum_{k=1}^{n} k = \frac{n(n+1)}{2} [/tex] In your example, you need the sum from 4 to 58, which is (58)(59)/2  (3)(4)/2 = 1705 So your total sum would be (1705)(35)  (55)(19) = 58630 



#7
Apr2610, 12:02 AM

P: 276

Awesome, that's exactly what I needed to hear! Thank you!
I see why 1705=k and how 35k fits into the first portion. Although, I'm not sure why you multiply the 19 in the original (35k19) by the # of steps when the first bit isn't multiplied by the # of steps (55). I see that you have found the right answer. Just trying to wrap my head around it. Thanks again! 



#8
Apr2610, 01:24 AM

P: 351

1705 is the sum of k for each step. I guess you could say that 19 is multiplied by one for each step, so:
[tex]\sum_{k=4}^{58} 1 = 55[/tex] 



#9
Apr2710, 09:59 PM

P: 276

Ahh! Thanks guys!
I greatly appreciate your help. The quick Q&A changed my perspective completely on the topic. 


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