Sequences and Series Problem. Help Pleeaassee

AI Thread Summary
The discussion revolves around solving a sequence and series problem involving the sum 2/(1*2*3) + 2/(2*3*4) + ... + 2/(100*101*102). The goal is to express the sum as a fraction in lowest terms, which is given as 2575/5151. Participants suggest using partial fraction decomposition to simplify the terms, emphasizing the importance of writing out the series to identify canceling terms. The key to solving the problem lies in recognizing how many terms will cancel out, leading to a manageable sum. Ultimately, the original poster finds clarity and resolves their confusion through the shared insights.
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Sequences and Series Problem. Help! Pleeaassee!

Homework Statement



I've attached the problem and my work. I'm supposed to express the sum as a fraction of numbers in lowest terms. The original statement was:
2/(1*2*3) + 2/(2*3*4) + 2/(3*4*5) + ... + 2/(100*101*102) and the answer is 2575/5151

In a similar problem, I solved it by decomposing it. So I tried the same thing with this problem, only I'm stuck because I don't know what to do to get to 2575/5151. Can someone please help me?
 

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I'm a little bit confused as to how you got your C value by plugging in n=1, you shoul have plugged in n=-2 (maybe you did and forgot to correct it because there were numbers crossed out). In any case, you have the correct partial fractions decomposition.

All you have to do now is write out the terms in the series and see what cancels out. A whole bunch of terms will cancel leaving you with a small sum that you have to calculate. I would suggest writing out the first 4 terms and the last 4 terms to see exactly which terms cancel out. You should be left with 4 numbers at the end that you sum together to get your answer!
 


The key is partial fraction decomposition:
<br /> \frac{1}{n (n + 1)(n + 2)} = \frac{A}{n} + \frac{B}{n + 1} + \frac{C}{n + 2}<br />

Multiply out to get rid of denominators, move everything on one side of the equation, collect like powers of n, and equate the obtained coefficients with zero. You will get thre linear equations for A, B, and C, that you need to solve.

After that, many neighboring terms in the sum will cancel, and you will be left with only the end terms.

EDIT:
I guess you did the hard work :smile:

Now, notice you may write the general term as:
<br /> \left(\frac{1}{n} - \frac{1}{n + 1} \right) - \left( \frac{1}{n + 1} - \frac{1}{n + 2} \right)<br />
Each term in the parentheses gives:
1/1 - 1/2 + 1/2 - 1/3 + ... + 1/n - !/(n + 1)
and
1/2 - 1/3 + 1/3 - 1/4 + ... + 1/(n + 1) - 1/(n + 2)

Do you see the cancelation?
 
Last edited:


Ohhhhhhhh I see. I've got it now, thank you guys soooooo much!
 
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