# Thread: (FP1) summation of finite series using standard results

1. ## (FP1) summation of finite series using standard results

How do I solve this some using the method of summation by standard results?

I am stuck on this for hours and hours, help would be appreciated big time.

2. $\sum\limits_{r=1}^{n}{\frac{1}{(r+3)(r+6)}}=\frac{ 1}{3}\sum\limits_{r=1}^{n}{\frac{(r+6)-(r+3)}{(r+3)(r+6)}},$ now the sum equals,

$\frac{1}{3}\sum\limits_{r=1}^{n}{\left( \frac{1}{r+3}-\frac{1}{r+6}+\frac{1}{r+4}-\frac{1}{r+4}+\frac{1}{r+5}-\frac{1}{r+5} \right)},$ which equals to,

$\frac{1}{3}\left\{ \sum\limits_{r=1}^{n}{\left( \frac{1}{r+3}-\frac{1}{r+4} \right)}+\sum\limits_{r=1}^{n}{\left( \frac{1}{r+4}-\frac{1}{r+5} \right)}+\sum\limits_{r=1}^{n}{\left( \frac{1}{r+5}-\frac{1}{r+6} \right)} \right\},$ and we happily telescope this to,

$\frac{1}{3}\left\{ \frac{1}{4}-\frac{1}{n+4}+\frac{1}{5}-\frac{1}{n+5}+\frac{1}{6}-\frac{1}{n+6} \right\}.$

3. Originally Posted by Krizalid
$\sum\limits_{r=1}^{n}{\frac{1}{(r+3)(r+6)}}=\frac{ 1}{3}\sum\limits_{r=1}^{n}{\frac{(r+6)-(r+3)}{(r+3)(r+6)}},$ now the sum equals,

$\frac{1}{3}\sum\limits_{r=1}^{n}{\left( \frac{1}{r+3}-\frac{1}{r+6}+\frac{1}{r+4}-\frac{1}{r+4}+\frac{1}{r+5}-\frac{1}{r+5} \right)},$ which equals to,

$\frac{1}{3}\left\{ \sum\limits_{r=1}^{n}{\left( \frac{1}{r+3}-\frac{1}{r+4} \right)}+\sum\limits_{r=1}^{n}{\left( \frac{1}{r+4}-\frac{1}{r+5} \right)}+\sum\limits_{r=1}^{n}{\left( \frac{1}{r+5}-\frac{1}{r+6} \right)} \right\},$ and we happily telescope this to,

$\frac{1}{3}\left\{ \frac{1}{4}-\frac{1}{n+4}+\frac{1}{5}-\frac{1}{n+5}+\frac{1}{6}-\frac{1}{n+6} \right\}.$
Thanks for going through the huge amount of latex.
One question though, can it be done by using only the standard results of $\sum\limits_{r=1}^{n}{r}, \sum\limits_{r=1}^{n}{r^2}, \sum\limits_{r=1}^{n}{r^3}$, as I am only a high school student and is only adept at this method . Probably I am required to do this sum by the method of standard results, too. So please give another try

4. Originally Posted by Krizalid
$\sum\limits_{r=1}^{n}{\frac{1}{(r+3)(r+6)}}=\frac{ 1}{3}\sum\limits_{r=1}^{n}{\frac{(r+6)-(r+3)}{(r+3)(r+6)}},$ now the sum equals,

$\frac{1}{3}\sum\limits_{r=1}^{n}{\left( \frac{1}{r+3}-\frac{1}{r+6}+\frac{1}{r+4}-\frac{1}{r+4}+\frac{1}{r+5}-\frac{1}{r+5} \right)},$ which equals to,

$\frac{1}{3}\left\{ \sum\limits_{r=1}^{n}{\left( \frac{1}{r+3}-\frac{1}{r+4} \right)}+\sum\limits_{r=1}^{n}{\left( \frac{1}{r+4}-\frac{1}{r+5} \right)}+\sum\limits_{r=1}^{n}{\left( \frac{1}{r+5}-\frac{1}{r+6} \right)} \right\},$ and we happily telescope this to,

$\frac{1}{3}\left\{ \frac{1}{4}-\frac{1}{n+4}+\frac{1}{5}-\frac{1}{n+5}+\frac{1}{6}-\frac{1}{n+6} \right\}.$
In the meantime I am trying to understand your working.

5. Have you ever seen what a telescoping sum is? It's not hard concept to diggest, so I suggest you to take a look.

As for your question, we can't apply those standard results, and the solution I gave you, it's the standard one. (I can offer you another one, but involves, calculus, and you don't want to see it.)

6. Originally Posted by Krizalid
Have you ever seen what a telescoping sum is? It's not hard concept to diggest, so I suggest you to take a look.

As for your question, we can't apply those standard results, and the solution I gave you, it's the standard one. (I can offer you another one, but involves, calculus, and you don't want to see it.)
I understand the concept, but I cannot do it myself, that's the catch
Guess I have to gear up, if your method is the only one viable.
Can I give a reputation to you for this?

7. Originally Posted by Krizalid
Have you ever seen what a telescoping sum is? It's not hard concept to diggest, so I suggest you to take a look.

As for your question, we can't apply those standard results, and the solution I gave you, it's the standard one. (I can offer you another one, but involves, calculus, and you don't want to see it.)
Another question, how do I get to know if you have quoted me? Except by visiting my thread of course.

Can I give a reputation to you for this?
It's voluntary. I can't force ya to give me reputation.

Another question, how do I get to know if you have quoted me? Except by visiting my thread of course.
Because you're the only one person I'm talkin' to... mmm what do you mean exactly?

9. Originally Posted by Krizalid
It's voluntary. I can't force ya to give me reputation.

Because you're the only one person I'm talkin' to... mmm what do you mean exactly?
No, in some forums members are not allowed to give reputations until they themselves have acquired some level of reputation, that is why I am asking.
Suppose I have posted in this forum and for some reason did not log in it for a few days afterwards, and my thread is lost under recent threads, how do I know whether someone had posted a reply or not?

As for your first sentence, well, this is not the case, you can give reputation anytime you want. (Actually, when pressing the "thanks" button, you're givin' reputation to that person, dunno if you realized it.)

11. Originally Posted by Krizalid

As for your first sentence, well, this is not the case, you can give reputation anytime you want. (Actually, when pressing the "thanks" button, you're givin' reputation to that person, dunno if you realized it.)
Thanks, then.