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,$$n.{ 2 }^{ n-1 }\left( x+\cfrac { { x }^{ 2 } }{ 2 } +\cfrac { { x }^{ 3 } }{ 6 } \right) +\cfrac { n\left( n-1 \right) }{ 2! } { 2 }^{ n-3 }{ \left( x+\cfrac { { x }^{ 2 } }{ 2 } \right) }^{ 2 }+\cfrac { n\left( n-1 \right) \left( n-2 \right) }{ 3! } { 2 }^{ n-3 }.{ x }^{ 3 }$$$$\therefore \cfrac { 2S }{ 6 } =\cfrac { n.{ 2 }^{ n-1 } }{ 6 } +\cfrac { n\left( n-1 \right) }{ 2! } { 2 }^{ n-2 }+\cfrac { n\left( n-1 \right) \left( n-2 \right) }{ 3! } { 2 }^{ n-3 }$$$$S={ n }^{ 2 }\left( n+3 \right) .{ 2 }^{ n-4 }$$Maths

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