# Discrete math set theory sum problem!

1. Mar 15, 2012

### charmedbeauty

1. The problem statement, all variables and given/known data

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)}

2. Relevant equations

3. 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 im really stuck I have no clue what to do... please help! Thanks

2. Mar 15, 2012

### 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

3. Mar 15, 2012

### 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?

4. Mar 15, 2012

### charmedbeauty

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

5. Mar 15, 2012

### scurty

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. Mar 15, 2012

### charmedbeauty

its (3k + 2) that needs to be simplified. the (9k +3) refers to the first part of the question! so somehow these are related im guessing (obviously its a factor of 3 although i dont 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. Mar 15, 2012

### tiny-tim

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

8. Mar 16, 2012

### charmedbeauty

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. Mar 16, 2012

### scurty

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!

10. Mar 16, 2012

### charmedbeauty

Last edited: Mar 16, 2012
11. Mar 16, 2012

### charmedbeauty

Is the last expression not the answer?

12. Mar 16, 2012

### 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.

13. Mar 16, 2012

### charmedbeauty

hmm just did it with a calculator still getting 55/6

14. Mar 16, 2012

### scurty

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

15. Mar 16, 2012

### charmedbeauty

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

16. Mar 17, 2012

### charmedbeauty

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

17. Mar 17, 2012

### scurty

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

18. Mar 17, 2012

### charmedbeauty

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