Discrete math set theory sum problem

[charmedbeauty]

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

 

[tiny-tim]

hi charmedbeauty!

write the LHS sums separately …

(and use some spaces! )

∑ 5/(k-1) - ∑ 3/k - ∑ 2/(k+2)​

and then change the variable in two of the sums

 

[scurty]

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 $\displaystyle\sum_{k=2}^{n} \frac{9k+6}{(n-1) \cdot n \cdot (n+2)}$, correct?

 

[charmedbeauty]

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 $\displaystyle\sum_{k=2}^{n} \frac{9k+6}{(n-1) \cdot n \cdot (n+2)}$, correct?

Thanks, but no... all the terms in the denominator are in terms of k and not n!

 

[scurty]

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?

 

[charmedbeauty]

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!

 

[tiny-tim]

hi charmedbeauty!

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)

 

[charmedbeauty]

hi charmedbeauty!


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

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.

 

[scurty]

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.

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

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

 

[charmedbeauty]

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!

 

[charmedbeauty]

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.

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

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

Is the last expression not the answer?

 

[scurty]

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.

 

[charmedbeauty]

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

 

[scurty]

The first terms that don't cancel are: $\frac{1}{3} \cdot \left(\frac{5}{1} - \frac{3}{2} + \frac{5}{2} - \frac{3}{3} + \frac{5}{3}\right) = \frac{20}{9}$

 

[charmedbeauty]

The first terms that don't cancel are: $\frac{1}{3} \cdot \left(\frac{5}{1} - \frac{3}{2} + \frac{5}{2} - \frac{3}{3} + \frac{5}{3}\right) = \frac{20}{9}$

Ohhh I thought the terms with three in the numerator cancelled.

 

[charmedbeauty]

Ohhh I thought the terms with three in the numerator cancelled.

Doesnt the 3/3 cancel out with the 5/5.?

 

[scurty]

I didn't get any $\frac{5}{5}$ term when I carried the partial sums out.

 

[charmedbeauty]

I didn't get any $\frac{5}{5}$ 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