Official solution
Given,
\[1 + {1 \over {1 + 2}} + {1 \over {1 + 2 + 3}}\, + \,....\,\, + \,\,{1 \over {1 + 2 + 3 + \,\,....\,\, + \,11}}\]
General term,
\[{T_n} = {1 \over {1 + 2 + 3\, + \,\,....\,\, + \,\,n}}\]
\[ = {1 \over {{{n(n + 1)} \over 2}}}\]
\[ = {2 \over {n(n + 1)}}\]
\[ = 2\left[ {{{n + 1 - n} \over {n(n + 1)}}} \right]\]
\[ = 2\left[ {{1 \over n} - {1 \over {n + 1}}} \right]\]
\[{t_1} = 2\left[ {{1 \over 1} - {1 \over 2}} \right]\]
\[{t_2} = 2\left[ {{1 \over 2} - {1 \over 3}} \right]\]
\[{t_3} = 2\left[ {{1 \over 3} - {1 \over 4}} \right]\]
\[ \vdots \]
\[{t_n} = 2\left[ {{1 \over n} - {1 \over {n + 1}}} \right]\]
\[\therefore\] \[{S_n} = {t_1} + {t_2} + {t_3}\, + \,\,....\,\, + \,\,{t_n}\]
\[ = 2\left[ {{1 \over 1} - {1 \over 2} + {1 \over 2} - {1 \over 3} + {1 \over 3} - {1 \over 4}\, + \,\,...\,\, + \,\,{1 \over n} - {1 \over {n + 1}}} \right]\]
\[ = 2\left[ {1 - {1 \over {n + 1}}} \right]\]
\[ = 2\left[ {{{n + 1 - 1} \over {n + 1}}} \right]\]
\[ = {{2n} \over {n + 1}}\]
\[\therefore\] \[{S_{11}} = {{2 \times 11} \over {11 + 1}} = {{22} \over {12}} = {{11} \over 6}\]