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Question

If n is a positive integer greater than $$3$$, show that $$n^3+\displaystyle\frac{n(n-1)}{\left\lfloor 2\right.}(n-2)^3+\frac{n(n-1)(n-2)(n-3)}{\left\lfloor 4\right.}(n-4)^3+....=n^2(n+3)2^{n-4}$$.


Solution

$$\Rightarrow { \left( { e }^{ x }+1 \right)  }^{ n }={ e }^{ nx }+{ c }_{ 1 }{ e }^{ \left( n-1 \right) x }+{ c }_{ 2 }{ e }^{ \left( n-2 \right) x }+....$$
$${ \left( { e }^{ x }-1 \right)  }^{ n }={ e }^{ nx }-{ c }_{ 1 }{ e }^{ \left( n-1 \right) x }+{ c }_{ 2 }{ e }^{ \left( n-2 \right) x }+....$$
$$2\left\{ { e }^{ nx }+{ c }_{ 2 }{ e }^{ \left( n-2 \right) x }+{ c }_{ 4 }{ e }^{ \left( n-4 \right) x }+.... \right\} $$
$$={ \left( { e }^{ x }+1 \right)  }^{ n }+{ \left( { e }^{ x }-1 \right)  }^{ n }$$
Equating co efficients of $${ x }^{ 3 }$$, we have -
$$\cfrac { 2 }{ 3! } \left\{ { n }^{ 3 }+\cfrac { n\left( n-1 \right)  }{ 2! } { \left( n-3 \right)  }^{ 3 }+... \right\} =$$ co efficient of $${ x }^{ 3 }$$ in $${ \left( { e }^{ x }+1 \right)  }^{ n }+{ \left( { e }^{ x }-1 \right)  }^{ n }$$
$$\Rightarrow \cfrac { 2 }{ 6 } S=$$ co efficient of $${ x }^{ 3 }$$ in $${ \left( 2+x+\cfrac { { x }^{ 2 } }{ 2 } +\cfrac { { x }^{ 3 } }{ 6 } +.... \right)  }^{ n }$$
i.e., in,
$$n.{ 2 }^{ n-1 }\left( x+\cfrac { { x }^{ 2 } }{ 2 } +\cfrac { { x }^{ 3 } }{ 6 }  \right) +\cfrac { n\left( n-1 \right)  }{ 2! } { 2 }^{ n-3 }{ \left( x+\cfrac { { x }^{ 2 } }{ 2 }  \right)  }^{ 2 }+\cfrac { n\left( n-1 \right) \left( n-2 \right)  }{ 3! } { 2 }^{ n-3 }.{ x }^{ 3 }$$
$$\therefore \cfrac { 2S }{ 6 } =\cfrac { n.{ 2 }^{ n-1 } }{ 6 } +\cfrac { n\left( n-1 \right)  }{ 2! } { 2 }^{ n-2 }+\cfrac { n\left( n-1 \right) \left( n-2 \right)  }{ 3! } { 2 }^{ n-3 }$$
$$S={ n }^{ 2 }\left( n+3 \right) .{ 2 }^{ n-4 }$$

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