Do you understand [b]why[/b] it is called a [i]telescoping[/i] series? Imagine an old fashioned telescope like a pirate might use.

Good night!

[tex]S=\frac{1}{1.4}+\frac{1}{4.7}+\cdots+\frac{1}{(3n-2)(3n+1)}[/tex]

Because:
[tex]a_1=1\\
a_2=4\\
r=a_2-a_1=4-1=3\\
a_n=a_1+(n-1)r\\
a_n=1+(n-1)3=1+3n-3\\
a_n=3n-2[/tex]

Now:
[tex]\frac{1}{(3n-2)(3n+1)}=\frac{a}{3n-2}+\frac{b}{3n+1}\\
a=\frac{1}{3}\\
b=-\frac{1}{3}[/tex]

So, the terms:

[tex]S=\frac{1}{1.4}+\frac{1}{4.7}+\cdots+\frac{1}{2998\cdot 3001}=\frac{1}{3.1}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{3.7}+\cdots+\frac{1}{3.2998}-\frac{1}{3.3001}[/tex]

[tex]S=\frac{1}{3}-\frac{1}{3.3001}=\frac{1000}{3001}[/tex]

Solved!

Hello.
I was studying the topic of arithmtic and progression series when I stumbled upon a new topic that is Telescoping series and It presented me with this problem:

1) Calculate the value of the following sum:[attachment=1].as.png[/attachment]
The answer is S=1000/3001

The notes I have are very vague about how to solve this , I really do not know where to start or what to do. It tells you in an example''[attachment=0]das.png[/attachment]'' that I can use the formula S=1/n+1 But it doesn't seem to be working in this case.
Could You provide some insight please?
How do I find The general term and how do I solve this?
Your help Is deeply appreciated thank you.
Thank You.
How do I find The general term and how do I solve this