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Question

If $$C$$ denotes the binomial coefficient $${ _{  }^{ n }{ C } }_{ r }$$ then $$(-1){ C }_{ 0 }^{ 2 }+2{ C }_{ 1 }^{ 2 }+5{ C }_{ 2 }^{ 2 }+....+(3n-1){ C }_{ n }^{ 2 }=$$


A
(3n2)2nCn
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B
(3n22)2nCn
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C
(5+3n)2nCn
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D
(3n52)2nCn
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Solution

The correct option is D $$\left( \cfrac { 3n-2 }{ 2 } \right) { _{ }^{ 2n }{ C } }_{ n }\quad $$
$${ \left( 1+x \right)  }^{ n }=^{ n }{ { C }_{ 0 } }+^{ n }{ { C }_{ 1 } }x+^{ n }{ { C }_{ 2 } }{ x }^{ 2 }+...+^{ n }{ { C }_{ n } }{ x }^{ n }$$
$${ \left( 1+\dfrac { 1 }{ x }  \right)  }^{ n }=^{ n }{ { C }_{ 0 } }+\dfrac { ^{ n }{ { C }_{ 1 } } }{ x } +\dfrac { ^{ n }{ { C }_{ 2 } } }{ { x }^{ 2 } } +...+\dfrac { ^{ n }{ { C }_{ n } } }{ { x }^{ n } } $$
$${ \left( 1+x \right)  }^{ n }{ \left( 1+\dfrac { 1 }{ x }  \right)  }^{ n }=\left( ^{ n }{ { C }_{ 0 } }+^{ n }{ { C }_{ 1 } }x+^{ n }{ { C }_{ 2 } }{ x }^{ 2 }+...+^{ n }{ { C }_{ n } }{ x }^{ n } \right) \left( ^{ n }{ { C }_{ 0 } }+\dfrac { ^{ n }{ { C }_{ 1 } } }{ x } +\dfrac { ^{ n }{ { C }_{ 2 } } }{ { x }^{ 2 } } +...+\dfrac { ^{ n }{ { C }_{ n } } }{ { x }^{ n } }  \right) $$
In RHS $${ ^{ n }{ { C }_{ 0 } } }^{ 2 }+{ ^{ n }{ { C }_{ 1 } } }^{ 2 }+...+{ ^{ n }{ { C }_{ n } } }^{ 2 }$$ comes as constant term.
So, $${ ^{ n }{ { C }_{ 0 } } }^{ 2 }+{ ^{ n }{ { C }_{ 1 } } }^{ 2 }+...+{ ^{ n }{ { C }_{ n } } }^{ 2 }$$ is constant term in $${ \left( 1+x \right)  }{ \left( 1+\dfrac { 1 }{ x }  \right)  }^{ n }$$
$$={ \dfrac { { \left( 1+x \right)  }^{ 2n } }{ { x }^{ n } }  }$$ 
The constant term in $${ \dfrac { { \left( 1+x \right)  }^{ 2n } }{ { x }^{ n } }  }$$ is $${ ^{ 2n }{ { C }_{ n } } }$$ as $${ \dfrac { { x }^{ n } }{ { x }^{ n } }  }$$ gets cancelled.
So, $${ ^{ n }{ { C }_{ 0 } } }^{ 2 }+{ ^{ n }{ { C }_{ 1 } } }^{ 2 }+...+{ ^{ n }{ { C }_{ n } } }^{ 2 }=^{ 2n }{ { C }_{ n } }$$
$${ { \left( 1+x \right)  }^{ n }=^{ n }{ { C }_{ 0 } }+^{ n }{ { C }_{ 1 } }x+^{ n }{ { C }_{ 2 } }{ x }^{ 2 }+...+^{ n }{ { C }_{ n } }{ x }^{ n } }$$
by differentiating 
$$n{ { \left( 1+x \right)  }^{ n-1 }=0.^{ n }{ { C }_{ 0 } }+1.^{ n }{ { C }_{ 1 } }+2.^{ n }{ { C }_{ 2 } }{ x }+...+n.^{ n }{ { C }_{ n } }{ x }^{ n-1 } }$$
$${ \left( 1+\dfrac { 1 }{ x }  \right)  }^{ n }=^{ n }{ { C }_{ 0 } }+\dfrac { ^{ n }{ { C }_{ 1 } } }{ x } +\dfrac { ^{ n }{ { C }_{ 2 } } }{ { x }^{ 2 } } +...+\dfrac { ^{ n }{ { C }_{ n } } }{ { x }^{ n } } $$
By multiplying 
$$n{ { \left( 1+x \right)  }^{ n-1 } }\times { \left( 1+\dfrac { 1 }{ x }  \right)  }^{ n }=\left( .^{ n }{ { C }_{ 0 } }+1.^{ n }{ { C }_{ 1 } }+2.^{ n }{ { C }_{ 2 } }{ x }+...+n^{ n }{ { C }_{ n } }{ x }^{ n-1 } \right) \times \left( ^{ n }{ { C }_{ 0 } }+\dfrac { ^{ n }{ { C }_{ 1 } } }{ x } +\dfrac { ^{ n }{ { C }_{ 2 } } }{ { x }^{ 2 } } +...+\dfrac { ^{ n }{ { C }_{ n } } }{ { x }^{ n } }  \right) $$
So, $$0.{ ^{ n }{ { C }_{ 0 } } }^{ 2 }+1.{ ^{ n }{ { C }_{ 1 } } }^{ 2 }+2.{ ^{ n }{ { C }_{ 2 } } }^{ 2 }+...+n.{ ^{ n }{ { C }_{ n } } }^{ 2 }$$ is coefficient of $$\dfrac { 1 }{ n } $$ in RHS
So, it is $$\dfrac { 1 }{ n } $$ term in $$n{ { \left( 1+x \right)  }^{ n-1 } }\times \dfrac { { \left( 1+x \right)  }^{ n } }{ { x }^{ n } } $$
i.e. $$n.^{ 2n-1 }{ { C }_{ n-1 } }=\dfrac { n }{ 2 } .^{ 2n }{ { C }_{ n } }$$
So, $$\sum { { (3n-1){ { C }_{ n } } }^{ 2 } } =3\sum { { n{ { C }_{ n } } }^{ 2 } } -\sum { { { { C }_{ n } } }^{ 2 } } $$
$$=\dfrac { 3n }{ 2 } .^{ 2n }{ { C }_{ n } }-^{ 2n }{ { C }_{ n } }=\dfrac { 3n-2 }{ 2 } \times  ^{ 2n }{ { C }_{ n } }$$



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