would you mind explaining that again. I dont get how you get that?

Thanks

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- May 21st 2008, 10:10 AMtaurus
- May 21st 2008, 10:15 AMKrizalid
Following up on TheEmptySet's post.

We have $\displaystyle \sum\limits_{k\,=\,1}^{n}{\left( a_{k+1}-a_{k} \right)}=\sum\limits_{k\,=\,1}^{n}{a_{k+1}}-\sum\limits_{k\,=\,1}^{n}{a_{k}}.$

Since

$\displaystyle \begin{aligned}

\sum\limits_{k\,=\,1}^{n}{a_{k+1}}&=\sum\limits_{k \,=\,2}^{n+1}{a_{k}} \\

& =\left( \sum\limits_{k\,=\,2}^{n}{a_{k}} \right)+a_{n+1} \\

& =\left( \sum\limits_{k\,=\,1}^{n}{a_{k}} \right)-a_{1}+a_{n+1},

\end{aligned}$

back substitute and $\displaystyle \sum\limits_{k\,=\,1}^{n}{\left( a_{k+1}-a_{k} \right)}=\left( \sum\limits_{k\,=\,1}^{n}{a_{k}} \right)-a_{1}+a_{n+1}-\left( \sum\limits_{k\,=\,1}^{n}{a_{k}} \right)=a_{n+1}-a_{1}.$

This result is called as "telescoping sum." - May 21st 2008, 10:19 AMMoo
Do you agree that $\displaystyle n^2+9n+20=(n+4)(n+5)$ ?

Now, let's find out A and B such that :

$\displaystyle \frac{-1}{(n+4)(n+5)}=\frac{A}{n+4}+\frac{B}{n+5}$

$\displaystyle \frac{A}{n+4}+\frac{B}{n+5}=\frac{A(n+5)+B(n+4)}{( n+4)(n+5)}=\frac{An+5A+Bn+4A}{(n+4)(n+5)}$

So we want $\displaystyle An+5A+Bn+4B=-1$

$\displaystyle n(A+B)+(5A+4B)=-1+0n$

-> $\displaystyle A+B=0 \longleftrightarrow A=-B$

And $\displaystyle 5A+4B=-1 \implies 5(-B)+4B=-1 \implies \boxed{B=1} \implies \boxed{A=-1}$

So $\displaystyle \frac{-1}{(n+4)(n+5)}=\frac{-1}{n+4}+\frac{1}{n+5}$

This may be better doing this way for you.. - May 21st 2008, 10:22 AMtaurus
last question

Moo

on that new way you showed me how would i get the sum?

i stick values for n of 0, 1, 2, 3... and see if there is number not reapeating? - May 21st 2008, 10:23 AMMoo
- May 21st 2008, 10:27 AMtaurus
but for the new method seems more understandable

i put n = 1 and got -1/30 which aint the sum:? - May 21st 2008, 10:32 AMMoo
It's n=1 in $\displaystyle \frac{-1}{n+4}$ :)

- May 21st 2008, 10:41 AMtaurus
but you said

(-1/n+4) + (1/n+5)

why dont you add? - May 21st 2008, 10:42 AMMoo
Because 1/(n+5) annulates with the following term, and so on...

Just write the few first terms of the series on a paper, and you will see that..