# Sequences and Series Problem. Help Pleeaassee

1. Mar 26, 2012

### theintarnets

Sequences and Series Problem. Help!!!! Pleeaassee!!!

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

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?

#### Attached Files:

• ###### math problem.png
File size:
44.7 KB
Views:
92
2. Mar 26, 2012

### scurty

Re: Sequences and Series Problem. Help!!!! Pleeaassee!!!

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!

3. Mar 26, 2012

### Dickfore

Re: Sequences and Series Problem. Help!!!! Pleeaassee!!!

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

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

Now, notice you may write the general term as:
$$\left(\frac{1}{n} - \frac{1}{n + 1} \right) - \left( \frac{1}{n + 1} - \frac{1}{n + 2} \right)$$
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: Mar 26, 2012
4. Mar 26, 2012

### theintarnets

Re: Sequences and Series Problem. Help!!!! Pleeaassee!!!

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