Help in inequality Please! :) (1 Viewer)

2013_jonathan_s · Jun 8, 2015

Dear all, can you please help me in this inequality proving?

leehuan · Jun 8, 2015

Can we please have an expert here? The (2n-k)! is throwing me off.

EDIT: It's giving me a permutation...

$Expansion\quad of\quad required\quad 'binomial':\\ \sum _{ k=0 }^{ n }{ { 2 }^{ k }\left( 2n-k \right) !\left( \begin{matrix} n \\ k \end{matrix} \right) { x }^{ k } } \\ =\left( 2n \right) !\left( \begin{matrix} n \\ 0 \end{matrix} \right) +2\left( 2n-1 \right) !\left( \begin{matrix} n \\ 1 \end{matrix} \right) x+{ 2 }^{ 2 }\left( 2n-2 \right) !\left( \begin{matrix} n \\ 2 \end{matrix} \right) { x }^{ 2 }+...+{ 2 }^{ n }\left( n \right) !\left( \begin{matrix} n \\ n \end{matrix} \right) { x }^{ n }$
$=n!\left[ \frac { \left( 2n \right) ! }{ n! } \left( \begin{matrix} n \\ 0 \end{matrix} \right) +2\frac { \left( 2n-1 \right) ! }{ n! } \left( \begin{matrix} n \\ 1 \end{matrix} \right) x+{ 2 }^{ 2 }\frac { \left( 2n-2 \right) ! }{ n! } \left( \begin{matrix} n \\ 2 \end{matrix} \right) { x }^{ 2 }+...+{ 2 }^{ n }\frac { n! }{ n! } \left( \begin{matrix} n \\ n \end{matrix} \right) { x }^{ n } \right] \\ =n!\left[ { _{ 2n }{ P }_{ n } }\left( \begin{matrix} n \\ 0 \end{matrix} \right) +2\cdot _{ 2n-1 }{ P }_{ n-1 }\left( \begin{matrix} n \\ 1 \end{matrix} \right) x+{ 2 }^{ 2 }\cdot _{ 2n-2 }{ P }_{ n-2 }\left( \begin{matrix} n \\ 2 \end{matrix} \right) { x }^{ 2 }+...+{ 2 }^{ n }\cdot _{ n }{ P }_{ 0 }\left( \begin{matrix} n \\ n \end{matrix} \right) { x }^{ n } \right]$

Help in inequality Please! :) (1 Viewer)

2013_jonathan_s

New Member

leehuan

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)