#2015_9 This is time for JEE Advanced..

We have ${ (1-x) }^{ n }={ C }_{ 0 }-{ C }_{ 1 }x+{ C }_{ 2 }{ x }^{ 2 }-{ C }_{ 3 }{ x }^{ 3 }+.....$

$\Rightarrow x{ (1-x) }^{ n }={ C }_{ 0 }x-{ C }_{ 1 }{ x }^{ 2 }+{ C }_{ 2 }{ x }^{ 3 }-{ C }_{ 3 }{ x }^{ 4 }+.....$

$\Rightarrow \int _{ 0 }^{ 1 }{ x{ (1-x) }^{ n } } dx=\int _{ 0 }^{ 1 }{ ( } { C }_{ 0 }x-{ C }_{ 1 }{ x }^{ 2 }+{ C }_{ 2 }{ x }^{ 3 }-{ C }_{ 3 }{ x }^{ 4 }+.....)dx\quad \quad \quad ----(1)$

LHS of $(1)$ = $\int _{ 0 }^{ 1 }{ x{ (1-x) }^{ n } } dx=\frac { 1 }{ (n+1)(n+2) } { \quad \quad \quad \left[ \because \int _{ 0 }^{ a }{ f\left( x \right) dx=\int _{ 0 }^{ a }{ f\left( a-x \right) } dx= } \left( \frac { { x }^{ n+1 } }{ n+1 } -\frac { { x }^{ n+2 } }{ n+2 } \right) \right] }_{ 0 }^{ 1 }$

RHS of $(1)$ = ${ \left[ \frac { { C }_{ 0 }{ x }^{ 2 } }{ 2 } -\frac { { C }_{ 1 }{ x }^{ 3 } }{ 3 } +\frac { { C }_{ 2 }{ x }^{ 4 } }{ 4 } -.... \right] }_{ 0 }^{ 1 }=\frac { { C }_{ 0 } }{ 2 } -\frac { { C }_{ 1 } }{ 3 } +\frac { { C }_{ 2 } }{ 4 } -.....$

Thus $\frac { { C }_{ 0 } }{ 2 } -\frac { { C }_{ 1 } }{ 3 } +\frac { { C }_{ 2 } }{ 4 } -\frac { { C }_{ 3 } }{ 5 } +...$ $=\boxed{ \frac { 1 }{ (n+1)(n+2) }}$