Answer
$\dfrac {1}{2}-\dfrac {1}{n+2}$
$ S_{1}=\dfrac {1}{6};$
$S_{2}=\dfrac {1}{4};$
$S_{3}=\dfrac {3}{10};$
$S_{4}=\dfrac {1}{3}$
Work Step by Step
$a_{n-1}+a_{n}=\dfrac {1}{\left( n-1\right) +1}-\dfrac {1}{\left( n-1\right) +2}+\dfrac {1}{n+1}-\dfrac {1}{n+2}=\dfrac {1}{n}-\dfrac {1}{n+2}$
$a_{n-2}+a_{n-1}+a_{n}=\dfrac {1}{\left( n-2\right) +1}-\dfrac {1}{\left( n-2\right) +2}+\dfrac {1}{\left( n-1\right) +1}-\dfrac {1}{\left( n-1\right) +2}+\dfrac {1}{n+1}-\dfrac {1}{n+2}=\dfrac {1}{n-1}-\dfrac {1}{n+2}$
$\Rightarrow a_{1}+a_{2}+\ldots a_{n-1}+a_{n}=\dfrac {1}{1+1}-\dfrac {1}{1+2}+\dfrac {1}{2+1}-\dfrac {1}{2+2}+\ldots +\dfrac {1}{\left( n-1\right) +1}-\dfrac {1}{\left( n-1\right) +2}+\dfrac {1}{n+1}-\dfrac {1}{n+2}=\dfrac {1}{1+1}-\dfrac {1}{n+2}=\dfrac {1}{2}-\dfrac {1}{n+2}$
$\Rightarrow S_{1}=\dfrac {1}{2\left( 1+2\right) }=\dfrac {1}{6};S_{2}=\dfrac {2}{2\left( 2+2\right) }=\dfrac {1}{4};S_{3}=\dfrac {3}{2\left( 3+2\right) }=\dfrac {3}{10};S_{4}=\dfrac {4}{2\left( 2+4\right) }=\dfrac {1}{3}$
