Method of difference (Series) 
Find $$\sum_{r=1}^{n}\frac{1}{r(r+3)}$$ by the method of differencing.

My attempt, 
$$\sum_{r=1}^{n}\frac{1}{r(r+3)}=\sum_{r=1}^{n}\left(\frac{1}{3r}-\frac{1}{3(r+3)}\right)$$
$$=\left(\frac{1}{3}-\frac{1}{12}\right)+\left(\frac{1}6-\frac{1}{15}\right)+\dots$$
This question is different as I cant see the common fraction. Hope someone can point it out.

$$\require{cancel}
\begin{align}
\sum_{r=1}^n\frac 1{r(r+3)}
&=\frac 13\sum_{r=1}^n\frac 1r-\frac 1{r+3}\\
&=\frac 13 \bigg[\bigg(\color{red}{\frac 11}-\cancel{\frac 14}\bigg)\\
&\;\;\quad+\bigg(\color{red}{\frac 12}-\cancel{\frac 15}\bigg)\\
&\;\;\quad+\bigg(\color{red}{\frac 13}-\cancel{\frac 16}\bigg)\\
&\;\;\quad+\bigg(\cancel{\frac 14}-\cancel{\frac 17}\bigg)\\
&\;\;\quad+\bigg(\cancel{\frac 15}-\cancel{\frac 18}\bigg)\\
&\qquad\qquad\vdots\\
&\;\;\quad+\bigg(\cancel{\frac 1{n-3}}-\cancel{\frac 1n}\bigg)\\
&\;\;\quad+\bigg(\cancel{\frac 1{n-2}}\color{blue}{-\frac 1{n+1}}\bigg)\\
&\;\;\quad+\bigg(\cancel{\frac 1{n-1}}\color{blue}{-\frac 1{n+2}}\bigg)\\
&\;\;\quad+\bigg(\cancel{\frac 1{n}}\color{blue}{-\frac 1{n+3}}\bigg)\bigg]\\
&=\frac 13\bigg[\color{red}{1+\frac 12+\frac 13}\color{blue}{-\bigg(\frac 1{n+1}+\frac 1{n+2}+\frac 1{n+3}}\bigg)\bigg]\\
&=\frac 13\bigg[\frac {11}6-\bigg(\frac 1{n+1}+\frac 1{n+2}+\frac 1{n+3}\bigg)\bigg]
\end{align}$$

Alternatively, in less graphical fashion, 
$$\begin{align}
\sum_{r=1}^n\frac 1{r(r+3)}
&=\frac 13\bigg[\sum_{r=1}^n\frac 1r-\sum_{r=1}^n\frac 1{r+3}\bigg]\\
&=\frac 13\bigg[\sum_{r=1}^n \frac 1r-\sum_{r=4}^{n+3}\frac 1r\bigg]\\
&=\frac 13\bigg[\bigg(\sum_{r=1}^3 \frac 1r+\sum_{r=4}^n \frac 1r\bigg)-\bigg(\sum_{r=4}^n \frac 1r+\sum_{r=n+1}^{n+3} \frac 1r\bigg)\bigg]\\
&=\frac 13\bigg[\sum_{r=1}^3 \frac 1r-\sum_{r=n+1}^{n+3}\frac 1r\bigg]\\
&=\frac 13\bigg[\frac {11}6-\bigg(\frac 1{n+1}+\frac 1{n+2}+\frac 1{n+3}\bigg)\bigg]
\end{align}$$
