1. Originally Posted by Moo
Hello

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

$\frac{-1}{(n+4)(n+5)}=\frac{n-n+4-5}{(n+4)(n+5)}=\frac{(n+4)-(n+5)}{(n+4)(n+5)}$
would you mind explaining that again. I dont get how you get that?

Thanks

2. Following up on TheEmptySet's post.

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

Since

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

This result is called as "telescoping sum."

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

Thanks
Do you agree that $n^2+9n+20=(n+4)(n+5)$ ?

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

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

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

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

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

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

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

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

This may be better doing this way for you..

4. 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?

5. 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

6. but for the new method seems more understandable
i put n = 1 and got -1/30 which aint the sum:?

7. It's n=1 in $\frac{-1}{n+4}$

8. but you said
(-1/n+4) + (1/n+5)

9. 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..

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