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."
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..