# Sum of series

Show 40 post(s) from this thread on one page
Page 2 of 2 First 12
• May 21st 2008, 10:10 AM
taurus
Quote:

Originally Posted by Moo
Hello :)

$\displaystyle n^2+9n+20=(n+4)(n+5)$

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

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

Thanks
• May 21st 2008, 10:15 AM
Krizalid
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 AM
Moo
Quote:

Originally Posted by taurus
would you mind explaining that again. I dont get how you get that?

Thanks

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 AM
taurus
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 AM
Moo
Quote:

Originally Posted by taurus
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?

The number which does not repeat is always the first one in a telescoping sum... I gave you a link to wikipedia and TheEmptySet explained it well.

And n begins at 1 (Wink)
• May 21st 2008, 10:27 AM
taurus
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 AM
Moo
It's n=1 in $\displaystyle \frac{-1}{n+4}$ :)
• May 21st 2008, 10:41 AM
taurus
but you said
(-1/n+4) + (1/n+5)