calculate-k-0-n-1-n-k-n-k-

Question Number 190788 by mnjuly1970 last updated on 11/Apr/23

calculate Ω= Σ_(k=0) ^n (( 1)/((n−k)!.(n+k )!))

$$ \\ $$$$\:\:\:\:\mathrm{calculate}\: \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\Omega=\:\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\frac{\:\mathrm{1}}{\left({n}−{k}\right)!.\left({n}+{k}\:\right)!} \\ $$$$ \\ $$

Answered by aleks041103 last updated on 12/Apr/23

(1/((n−k)!(n+k)!))=(1/((2n)!)) (((2n)!)/((n−k)!(2n−(n−k))!))= =(1/((2n)!)) (((2n)),((n−k)) ) ⇒Ω = (1/((2n)!))Σ_(k=0) ^n (((2n)),((n−k)) ) = = (1/((2n)!)) Σ_(k=0) ^n (((2n)),(k) ) = = (1/((2n)!)) (( (((2n)),(n) ) +Σ_(k=0) ^(2n) (((2n)),(k) ))/2)= =((2^(2n) +(((2n)!)/((n!)^2 )))/((2n)!)) ⇒Ω = (4^n /((2n)!)) + (1/((n!)^2 ))

$$\frac{\mathrm{1}}{\left({n}−{k}\right)!\left({n}+{k}\right)!}=\frac{\mathrm{1}}{\left(\mathrm{2}{n}\right)!}\:\frac{\left(\mathrm{2}{n}\right)!}{\left({n}−{k}\right)!\left(\mathrm{2}{n}−\left({n}−{k}\right)\right)!}= \\ $$$$=\frac{\mathrm{1}}{\left(\mathrm{2}{n}\right)!}\begin{pmatrix}{\mathrm{2}{n}}\\{{n}−{k}}\end{pmatrix} \\ $$$$\Rightarrow\Omega\:=\:\frac{\mathrm{1}}{\left(\mathrm{2}{n}\right)!}\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\begin{pmatrix}{\mathrm{2}{n}}\\{{n}−{k}}\end{pmatrix}\:= \\ $$$$=\:\frac{\mathrm{1}}{\left(\mathrm{2}{n}\right)!}\:\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\begin{pmatrix}{\mathrm{2}{n}}\\{{k}}\end{pmatrix}\:= \\ $$$$=\:\frac{\mathrm{1}}{\left(\mathrm{2}{n}\right)!}\:\frac{\begin{pmatrix}{\mathrm{2}{n}}\\{{n}}\end{pmatrix}\:+\underset{{k}=\mathrm{0}} {\overset{\mathrm{2}{n}} {\sum}}\begin{pmatrix}{\mathrm{2}{n}}\\{{k}}\end{pmatrix}}{\mathrm{2}}= \\ $$$$=\frac{\mathrm{2}^{\mathrm{2}{n}} +\frac{\left(\mathrm{2}{n}\right)!}{\left({n}!\right)^{\mathrm{2}} }}{\left(\mathrm{2}{n}\right)!} \\ $$$$ \\ $$$$\Rightarrow\Omega\:=\:\frac{\mathrm{4}^{{n}} }{\left(\mathrm{2}{n}\right)!}\:+\:\frac{\mathrm{1}}{\left({n}!\right)^{\mathrm{2}} } \\ $$

Facebook Tweet Pin

calculate-k-0-n-1-n-k-n-k-

Leave a Reply Cancel reply