\(\displaystyle \begin{align*}\sum\limits_{k = 1}^{18} {\frac{1}{{{x^2} + 3x + 2}}} &= \sum\limits_{k = 1}^{18} {\frac{1}{{x + 2}} - \frac{1}{{x + 1}}} \\&=\left(\dfrac{1}{3}-\dfrac{1}{2}\right)+\left(\dfrac{1}{4}-\dfrac{1}{3}\right)+\left(\dfrac{1}{5}-\dfrac{1}{4}\right)+\cdots +\left(\dfrac{1}{20}-\dfrac{1}{19}\right)\\&=\left(\dfrac{1}{20}-\dfrac{1}{2}\right) \end{align*}\)Yes but how do you apply that when it comes to a given function?
a0 is the 1st term. Note immediately that the subscript is 1 less than term number.What does ai, ai-1, an, and a0 mean? I studied this subject without getting too deep in telescoping sums because I don't understand this. Can anyone explain to me? Any help will be welcome, thank you.
Professor, I check three times to make sure that you 1st line was in fact wrong. It seems that you are off by a negative sign.\(\displaystyle \begin{align*}\sum\limits_{k = 1}^{18} {\frac{1}{{{x^2} + 3x + 2}}} &= \sum\limits_{k = 1}^{18} {\frac{1}{{x + 2}} - \frac{1}{{x + 1}}} \\&=\left(\dfrac{1}{3}-\dfrac{1}{2}\right)+\left(\dfrac{1}{4}-\dfrac{1}{3}\right)+\left(\dfrac{1}{5}-\dfrac{1}{4}\right)+\cdots +\left(\dfrac{1}{20}-\dfrac{1}{19}\right)\\&=\left(\dfrac{1}{20}-\dfrac{1}{2}\right) \end{align*}\)
Professor, I check three times to make sure that you 1st line was in fact wrong. It seems that you are off by a negative sign.
You are correct. I caught that also. But the new damn time limit prevented me from correcting it.Professor, I check three times to make sure that you 1st line was in fact wrong. It seems that you are off by a negative sign.
Now ... now ...You are correct. I caught that also. But the new damn time limit prevented me from correcting it.
It should be
\(\displaystyle \begin{align*}\sum\limits_{x = 1}^{18} {\left( {\frac{1}{{{x^2} + 3x + 2}}} \right)} &= \sum\limits_{x = 1}^{18} {\left( {\frac{1}{{x + 1}} - \frac{1}{{x + 2}}} \right)}\\&=\left( {\frac{1}{2} - \frac{1}{3}} \right) + \left( {\frac{1}{3} - \frac{1}{4}} \right) + \cdots + \left( {\frac{1}{{19}} - \frac{1}{{20}}} \right) \\&= \left( {\frac{1}{2} - \frac{1}{{20}}} \right)\end{align*}\)
pka is a retired research mathematician! That says it all (at least for me). If he wants to say damn, I give him permission. pka deserves whatever entitlements he wants in order to volunteer on this forum.Now ... now ...
IF it is a telescoping seriesSo I start from the beginning, I make sure cancelation, and those that remain (the first and the last one) is the solution. Is that correct?
No, he doesn't deserve "whatever entitlements" than anyone else. He's just another username on the board. Don't give attributions topka is a retired research mathematician! That says it all (at least for me). If he wants to say damn, I give him permission. pka deserves whatever entitlements he wants in order to volunteer on this forum.
nd remove his posts agai
You know what I say to that? ?lookagain, pka (and even yourself) is a great addition to this forum. pka can solve every problem posted here and do it correctly. That deserves pka some entitlement. I am just saying we should not say a word if pka uses a word that may be inappropriate. After all, all I am saying is to allow him to say a word, not very much in my opinion.
Now... now ... now (triple nows)