Discrete math set theory sum problem

In summary, In this problem, the first part asks to simplify an expression in terms of a single denominator, while the second part asks for an expression in terms of a single numerator. However, the first numerator is incorrect because it includes terms that are not in terms of k. The correct numerator is (9k+6), which is three times (3k+2). Both expressions are then related to one another because (9k+6) is equal to (3k+2).
  • #1
charmedbeauty
271
0

Homework Statement



Prove that if k>1 then,

5/(k-1)-3/k-2/(k+2) = (9k+6)/(k-1)k(k+2)

Hence simplify Ʃ of k=2 to n {(3k+2)/(k-1)k(k+2)}


Homework Equations





The Attempt at a Solution



Ok so the first part is ok I just multiplied the denominators with the numerators and expanded and cancelled.


However the second part (hence simplify)... is where I am really stuck I have no clue what to do... please help! Thanks
 
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  • #2
hi charmedbeauty! :smile:

write the LHS sums separately …

(and use some spaces! :rolleyes:)

∑ 5/(k-1) - ∑ 3/k - ∑ 2/(k+2)​

and then change the variable in two of the sums :wink:
 
  • #3
Alternatively, you can write out the first few (I would recommend at least 8) sums of the series (the series that Tim wrote out) and see if you can spot of pattern of terms canceling. Most telescoping series problems work out nicely with canceling; the answer for this one isn't pretty looking, but it is solvable!

And, just out of curiosity, you meant to type

Hence simplify [itex]\displaystyle\sum_{k=2}^{n} \frac{9k+6}{(n-1) \cdot n \cdot (n+2)}[/itex], correct?
 
  • #4
scurty said:
Alternatively, you can write out the first few (I would recommend at least 8) sums of the series (the series that Tim wrote out) and see if you can spot of pattern of terms canceling. Most telescoping series problems work out nicely with canceling; the answer for this one isn't pretty looking, but it is solvable!

And, just out of curiosity, you meant to type

Hence simplify [itex]\displaystyle\sum_{k=2}^{n} \frac{9k+6}{(n-1) \cdot n \cdot (n+2)}[/itex], correct?

Thanks, but no... all the terms in the denominator are in terms of k and not n!
 
  • #5
charmedbeauty said:
Thanks, but no... all the terms in the denominator are in terms of k and not n!

Oops! I meant for the terms in the denominator to be k! I was referring to your inconsistency between the 3k+2 in the numerator and the 9k+6 in the numerator. Which is it? Have you worked out a solution yet?
 
  • #6
scurty said:
Oops! I meant for the terms in the denominator to be k! I was referring to your inconsistency between the 3k+2 in the numerator and the 9k+6 in the numerator. Which is it? Have you worked out a solution yet?

its (3k + 2) that needs to be simplified. the (9k +3) refers to the first part of the question! so somehow these are related I am guessing (obviously its a factor of 3 although i don't know if this has any significance.)

But to clarify The first part of the question asks to simply the expression in terms of a single denominator you do this and you get (9k +3) as the numerator.
The next part has a DIFFERENT numerator, namely, (3k+2).

AND THIS IS EXACTLY WHERE I'M STUCK!
 
  • #7
hi charmedbeauty! :smile:
charmedbeauty said:
The first part of the question asks to simply the expression in terms of a single denominator you do this and you get (9k +3) as the numerator.
The next part has a DIFFERENT numerator, namely, (3k+2).

no, the first numerator was (9k+6), which is exactly 3 times (3k+2) :wink:
 
  • #8
tiny-tim said:
hi charmedbeauty! :smile:


no, the first numerator was (9k+6), which is exactly 3 times (3k+2) :wink:

Sorry, yes that is correct.

Anyway I have worked it out now ...

1/3(55/6-5/n-2/n+1-2/n+2)

thanks for the input.
 
  • #9
Your constant term is a little bit off, but otherwise the other terms of your answer are correct. You probably made a little error when adding up your constant terms, that's all.

Also, I know what you meant by your answer, but in the future use parentheses around the denominator of fractions so it avoids confusion.

[itex]1/3(55/6-5/n-2/n+1-2/n+2)= \displaystyle\frac{1}{3} \cdot \left(\frac{55}{6} - \frac{5}{n} - \frac{2}{n} + 1 - \frac{2}{n} + 2\right) \neq \frac{1}{3} \cdot \left(\frac{55}{6} - \frac{5}{n} - \frac{2}{n+1} - \frac{2}{n+2}\right)[/itex]

Like I mentioned earlier, the constant term is off, retry the calculation and see you messed up!
 
  • #10
scurty said:
Your constant term is a little bit off, but otherwise the other terms of your answer are correct. You probably made a little error when adding up your constant terms, that's all.

see next post down!
 
Last edited:
  • #11
scurty said:
Your constant term is a little bit off, but otherwise the other terms of your answer are correct. You probably made a little error when adding up your constant terms, that's all.

Also, I know what you meant by your answer, but in the future use parentheses around the denominator of fractions so it avoids confusion.

[itex]1/3(55/6-5/n-2/n+1-2/n+2)= \displaystyle\frac{1}{3} \cdot \left(\frac{55}{6} - \frac{5}{n} - \frac{2}{n} + 1 - \frac{2}{n} + 2\right) \neq \frac{1}{3} \cdot \left(\frac{55}{6} - \frac{5}{n} - \frac{2}{n+1} - \frac{2}{n+2}\right)[/itex]

Like I mentioned earlier, the constant term is off, retry the calculation and see you messed up!

Is the last expression not the answer?
 
  • #12
The constant term is a little bit off. The terms involving the n variables are correct. Just check to see if you added up the constant terms correctly.
 
  • #13
scurty said:
The constant term is a little bit off. The terms involving the n variables are correct. Just check to see if you added up the constant terms correctly.


hmm just did it with a calculator still getting 55/6
 
  • #14
The first terms that don't cancel are: [itex]\frac{1}{3} \cdot \left(\frac{5}{1} - \frac{3}{2} + \frac{5}{2} - \frac{3}{3} + \frac{5}{3}\right) = \frac{20}{9}[/itex]
 
  • #15
scurty said:
The first terms that don't cancel are: [itex]\frac{1}{3} \cdot \left(\frac{5}{1} - \frac{3}{2} + \frac{5}{2} - \frac{3}{3} + \frac{5}{3}\right) = \frac{20}{9}[/itex]

Ohhh I thought the terms with three in the numerator cancelled.
 
  • #16
charmedbeauty said:
Ohhh I thought the terms with three in the numerator cancelled.

Doesnt the 3/3 cancel out with the 5/5.?
 
  • #17
I didn't get any [itex]\frac{5}{5}[/itex] term when I carried the partial sums out.
 
  • #18
scurty said:
I didn't get any [itex]\frac{5}{5}[/itex] term when I carried the partial sums out.

No sorry my mistake I checked it again not realising that the 5/4 went with the -3/4 and -2/4
 

Related to Discrete math set theory sum problem

What is discrete math set theory?

Discrete math set theory is a branch of mathematics that deals with the study of finite or countable sets. It involves the study of operations on sets, such as union, intersection, and complement, as well as the properties of these operations.

What is a sum problem in discrete math set theory?

A sum problem in discrete math set theory involves finding the sum of elements in a set. This is typically done using the summation notation, where the elements of the set are added together using a specific formula or pattern.

How do you solve a sum problem in discrete math set theory?

To solve a sum problem in discrete math set theory, you first need to identify the formula or pattern that is used to add the elements in the set. Then, you can use this formula to calculate the sum of the elements in the set.

What are some common formulas used in solving sum problems in discrete math set theory?

Some common formulas used in solving sum problems in discrete math set theory include the arithmetic series formula, geometric series formula, and the binomial theorem. These formulas can be used to find the sum of elements in different types of sets.

What are some real-life applications of discrete math set theory?

Discrete math set theory has many real-life applications, such as in computer science, cryptography, and finance. It is also used in decision-making, data analysis, and problem-solving in various fields such as engineering, economics, and social sciences.

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