Extremely Difficult sequences question

[not-cody]

Hi guys!

Im new here so Im just hoping that i got the correct category!
So, i think im just supposed to give you guys the question and what i think of it.

Alright!
So we have sequence
1/(2*4) + 1/(4*6) + 1/(6*8).......1/(2008*2012) = ?
Also, what is the rule for it? (e.g: n^2 + 5)

I have no idea how to start this, but I would also appreciate a form of solution!

Thanks!

 

[stapel]

...we have sequence
1/(2*4) + 1/(4*6) + 1/(6*8).......1/(2008*2012) = ?
Also, what is the rule for it? (e.g: n^2 + 5)

I have no idea how to start this...

A good place to start might be in looking at what changes, from what term to the next, and looking for the pattern.

For instance, all the terms have 1 as the numerator, so that's probably part of the pattern. The first term, for n = 1, has the form 1/(2*4). The second term, for n = 2, has the form 1/(4*6). The only part that changed was the denominator. When n = 1, the first factor was 2, which can be stated as 2*1; the other factor was two greater, for 2*1 + 2. When n = 2, the first factor was 4, which can be stated as 2*2; the other factor was two greater, for 2*2 + 2. Is there a pattern here? Does that pattern work for the third term, when n = 3?

...and so forth.

 

[not-cody]

I think ive got it

From what you guys have said (Thank you so much by the way!!!) The rule that i was able to find is:

___1___
4n^2+4n

Which seems like the correct equation. But how am i supposed to apply it to the last part of the equation 1/2008*2010 (Yeah i made a mistake on that one, sorry ) ?
From what ive seen, it should be something like 1004 ?

Thanks I truly appreciate it. If i have gotten it right, no need to respond, Ill just thank!

 

[stapel]

The rule that i was able to find is:

___1___
4n^2+4n

...how am i supposed to apply it to the last part of the equation 1/2008*2010...?

What value of n gives you 2n = 2008 and 2n+2 = 2010?

What summation formulas have they given you? What partial-fraction decomposition did you find? What else have you tried?

Please be complete. Thank you!

 

[not-cody]

1004 because....

Ok, so i believe that it is 1004 because if you equate it:

4n^2 + 4n = 2008*2010

4n^2 + 4n = 4036080

4n^2 + 4n - 4036080 = 0

n = 1004, or n = -1005

right?

Now what i am worried about is if i attempt to substitute them into the formula, would i simply get the answer to the value (1/2008*2010) or would it be the entire sum ?

AAAAAARGGHHH this is driving me up the wall

Cheers!

 

[lookagain]

We have sequence

1/(2*4) + 1/(4*6) + 1/(6*8) + ... + 1/(2008*2010) = ?

I have no idea how to start this, but I would also appreciate a form of solution!

That's not a sequence. It's a series.




Here is the corresponding sequence for it (if it had been a different question):

1/(2*4), 1/(4*6), 1/(6*8), ... , 1/(2008*2010)

 

[lookagain]

I can tell you that the 1st 1004 terms add up to 251/1005 (reduced fraction).

That's from a computer program run...


The series is neither arithmetic or geometric, \(\displaystyle \ \ \ \ \) <------ That is true.

so say bye to a formula! \(\displaystyle \ \ \ \ \)No, that's not true.\(\displaystyle \ \ \ \)See below.

The formula is just \(\displaystyle \ \dfrac{n}{ \ 4(n + 1) \ }.\)



It's a consequence of the partial-fraction decomposition that stapel stated, which then leads to a
telescoping series where all terms cancel out, except for the first and the last terms.



not-cody, were you shown how to do partial-fraction decompositions?

 

[soroban]

Hello, not-cody!

We have a series: .\(\displaystyle S \;=\;\frac{1}{2\cdot4} + \frac{1}{4\cdot6} + \frac{1}{6\cdot8} + \cdots + \frac{1}{2008\cdot2010}\)
Find \(\displaystyle S.\)

This is what lookagain suggested.

Each fraction can be decomposed by Partial Fractions,.
. . \(\displaystyle \frac{1}{2n(2n+2)} \;=\;\frac{1}{2}\left[\frac{1}{2n} - \frac{1}{2n+2}\right]\)

Then the series becomes:
. . \(\displaystyle S \;=\;\frac{1}{2}\left[\left(\frac{1}{2} - \frac{1}{4}\right) + \left(\frac{1}{4} - \frac{1}{6}\right) + \left(\frac{1}{6} - \frac{1}{8}\right) + \cdots + \left(\frac{1}{2008} - \frac{1}{2010}\right)\right] \)

Nearly all the terms cancel out.

We are left with: .\(\displaystyle S \;=\;\frac{1}{2}\left[\frac{1}{2} - \frac{1}{2010}\right] \;=\;\frac{1}{2}\left(\frac{2008}{2(2010)}\right) \)

Therefore: .\(\displaystyle S \;=\;\dfrac{251}{1005}\)