Answer
$${S_1} = \frac{1}{6},{S_2} = \frac{1}{4},{S_3} = \frac{3}{{10}},{S_4} = \frac{1}{3},{S_5} = \frac{5}{{14}}$$
Work Step by Step
$$\eqalign{
& {a_n} = \frac{1}{{\left( {n + 1} \right)\left( {n + 2} \right)}} \cr
& {\text{Find the first five terms of the sequence}} \cr
& {a_1} = \frac{1}{{\left( {1 + 1} \right)\left( {1 + 2} \right)}} = \frac{1}{6} \cr
& {a_2} = \frac{1}{{\left( {2 + 1} \right)\left( {2 + 2} \right)}} = \frac{1}{{12}} \cr
& {a_3} = \frac{1}{{\left( {3 + 1} \right)\left( {3 + 2} \right)}} = \frac{1}{{20}} \cr
& {a_4} = \frac{1}{{\left( {4 + 1} \right)\left( {4 + 2} \right)}} = \frac{1}{{30}} \cr
& {a_5} = \frac{1}{{\left( {5 + 1} \right)\left( {5 + 2} \right)}} = \frac{1}{{42}} \cr
& {\text{Then by definition of partial sum}} \cr
& {S_1} = {a_1} = \frac{1}{6} \cr
& {S_2} = {a_1} + {a_2} = \frac{1}{6} + \frac{1}{{12}} = \frac{1}{4} \cr
& {S_3} = {a_1} + {a_2} + {a_3} = \frac{1}{6} + \frac{1}{{12}} + \frac{1}{{20}} = \frac{3}{{10}} \cr
& {S_4} = {a_1} + {a_2} + {a_3} + {a_4} = \frac{1}{6} + \frac{1}{{12}} + \frac{1}{{20}} + \frac{1}{{30}} = \frac{1}{3} \cr
& {S_5} = {a_1} + {a_2} + {a_3} + {a_4} + {a_5} = \frac{1}{6} + \frac{1}{{12}} + \frac{1}{{20}} + \frac{1}{{30}} + \frac{1}{{42}} = \frac{5}{{14}} \cr
& {\text{the partial sums are}}: \cr
& {S_1} = \frac{1}{6},{S_2} = \frac{1}{4},{S_3} = \frac{3}{{10}},{S_4} = \frac{1}{3},{S_5} = \frac{5}{{14}} \cr} $$
