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.
$\displaystyle \sum\limits_{r=1}^{n}{\frac{1}{(r+3)(r+6)}}=\frac{ 1}{3}\sum\limits_{r=1}^{n}{\frac{(r+6)-(r+3)}{(r+3)(r+6)}},$ now the sum equals,
$\displaystyle \frac{1}{3}\sum\limits_{r=1}^{n}{\left( \frac{1}{r+3}-\frac{1}{r+6}+\frac{1}{r+4}-\frac{1}{r+4}+\frac{1}{r+5}-\frac{1}{r+5} \right)},$ which equals to,
$\displaystyle \frac{1}{3}\left\{ \sum\limits_{r=1}^{n}{\left( \frac{1}{r+3}-\frac{1}{r+4} \right)}+\sum\limits_{r=1}^{n}{\left( \frac{1}{r+4}-\frac{1}{r+5} \right)}+\sum\limits_{r=1}^{n}{\left( \frac{1}{r+5}-\frac{1}{r+6} \right)} \right\},$ and we happily telescope this to,
$\displaystyle \frac{1}{3}\left\{ \frac{1}{4}-\frac{1}{n+4}+\frac{1}{5}-\frac{1}{n+5}+\frac{1}{6}-\frac{1}{n+6} \right\}.$
Thanks for going through the huge amount of latex.
One question though, can it be done by using only the standard results of $\displaystyle \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
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.)
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?
Two options: you should have a notification by mail that you thread was replied; see your profile and click on "statistics" and go to "Find all threads started by ssadi," that should lead you to all your threads.
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.)