Cubing unities?

First of all, $\cot ^{ -1 }{ \left( { i }^{ 2 }+i+1 \right) }$ can be written as $\tan ^{ -1 }{ \frac { 1 }{ { i }^{ 2 }+i+1 } }$ . So, ${ S }_{ n }=\sum _{ i=1 }^{ n }{ \tan ^{ -1 }{ \frac { 1 }{ { i }^{ 2 }+i+1 } } } \\ =\sum _{ i=1 }^{ n }{ \tan ^{ -1 }{ \frac { 1 }{ { i }(i+1)+1 } } } \\ =\sum _{ i=1 }^{ n }{ \tan ^{ -1 }{ \frac { (i+1)-(i) }{ { i }(i+1)+1 } } } \\ =\sum _{ i=1 }^{ n }{ \tan ^{ -1 }{ (i+1) } -\tan ^{ -1 }{ i } } \\ =\tan ^{ -1 }{ (n+1) } -\tan ^{ -1 }{ 1 } \\ =\tan ^{ -1 }{ \left( \frac { n }{ n+2 } \right) } \\ =\cot ^{ -1 }{ \left( \frac { n+2 }{ n } \right) }$ Hence, ${ S }_{ 100 }=\cot ^{ -1 }{ \left( \frac { 102 }{ 100 } \right) } =\cot ^{ -1 }{ \left( \frac { 51 }{ 50 } \right) } \\ So,\quad 2*(a+b)=2(101)=202$