JEE Problem

First solving the inner bracket,

$1+\sum _{ k=1 }^{ n }{ 2k }$ = 1+ $\frac { n }{ 2 } \left( 4+2(n-1) \right)$ = ${ n }^{ 2 }+n+1$

Now,

$\cot ^{ -1 }{ \left( { n }^{ 2 }+n+1 \right) =\cot ^{ -1 }{ \left( \frac { n\left( n+1 \right) +1 }{ (n+1)-n } \right) } }=\cot ^{ -1 }{ n } -\cot ^{ -1 }{ (n+1 } )$

Using $V_{n}$ summation,

$\sum _{ n=1 }^{ 23 }{ \left( \cot ^{ -1 }{ n } -\cot ^{ -1 }{ (n+1 } ) \right) } = \cot ^{ -1 }{ 1 } -\cot ^{ -1 }{ 24 }$

$\cot { \left( \cot ^{ -1 }{ 1-\cot ^{ -1 }{ 24 } } \right) } =\quad \cot { \left( \cot ^{ -1 }{ \left( \frac { 24+1 }{ 23 } \right) } \right) } = \frac { 25 }{ 23 }$

$\boxed{25-23=2}$