Inspired by Calvin Lin - Part II

${ T }_{ r+1 }={ \left\{ x+{ \left( -1 \right) }^{ r }r \right\} }^{ 2 }={ x }^{ 2 }+{ r }^{ 2 }+2x{ \left( -1 \right) }^{ r }r\\ \\ f\left( n \right) =\sum _{ r=0 }^{ r=2n }{ \left[ { x }^{ 2 }+{ r }^{ 2 }+2x{ \left( -1 \right) }^{ r }r \right] } \\ \\ =\left( 2n+1 \right) { x }^{ 2 }+\cfrac { \left( 2n \right) \left( 2n+1 \right) \left( 4n+1 \right) }{ 6 } +2xn\\ \\ f\left( n \right) \quad will\quad be\quad min\quad for\quad x=\cfrac { -n }{ 2n+1 } \\ \\ { f\left( n \right) }_{ min }=\cfrac { n\left( 2n+1 \right) \left( 4n+1 \right) }{ 3 } -\cfrac { { n }^{ 2 } }{ 2n+1 } =\cfrac { 3936170 }{ 59 } \\ \\ or,\quad n=29$

Note- How did I solve the eqn? When u simplify ,u will notice that f(n) is increasing for n>0 ,so putting 10,20,30... u will find range in which root is then put 25...27...29....yes hit & trial with a scientific calculator or use standard method for quadratic equation