Question

Cum ajung la solutie ?

Answer 1

   
[tex]\displaystyle\\ \text{Sa se determine numarul de elemente al multimii:}\\\\ E = \left\{ n\in Q~ \Big|~ \frac{A_{2n+2}^4}{(2n+2)!}\ \textless \ \frac{10}{(2n-1)!} \right\}\\\\ \text{Rezolvam inecuatia:}\\\\ \frac{A_{2n+2}^4}{(2n+2)!}\ \textless \ \frac{10}{(2n-1)!}\\\\ \frac{(2n+2)(2n+1)(2n)(2n-1)}{(2n-2)!\times (2n-1)(2n)(2n+1)(2n+2)}\ \textless \ \frac{10}{(2n-1)!}\\\\ \text{Simplificam cele 4 paranteze.}\\\\ \frac{1}{(2n-2)!}\ \textless \ \frac{10}{(2n-1)!}\\\\ \frac{(2n-1)!}{(2n-2)!}\ \textless \ 10\\\\ \frac{(2n-2)! \times (2n-1)}{(2n-2)!}\ \textless \ 10 [/tex]


[tex]\text{Simplificam factorialii.}\\ 2n-1\ \textless \ 10\\ 2n\ \textless \ 11\\ n\ \textless \ \frac{11}{2}\\ \boxed{\bf n\ \textless \ 5,5}\\ \text{Mai avem de rezolvat o inecuatie.}\\ \text{In enuntul initial avem la numaratorul primei factii, expresia:}\\ \boxed{A_{2n+2}^4 }\\ \text{Din aceasta expresie rezulta:}\\ 2n+2\geq4\\ 2n\geq4-2\\ 2n\geq2\\ n\geq\frac{2}{2} \\ \boxed{\bf n\geq1}\\ \Longrightarrow~~\boxed{\bf1\leq n\ \textless \ 5,5}\\ \Longrightarrow~~\boxed{\bf E =\{1;~1,5;~2;~2,5;~3;~3,5;~4;~4,5;~5\}}\\ [/tex]


[tex]\text{Multimea E are un numar de: }~\boxed{\bf9~\text{\bf elemente}} \\\\ \text{Precizare:}\\\\ \text{Sunt admise elementele: } 1,5;~2,5;~3,5;~4,5 ~\text{deoarece}\\ \text{n apare in aranjamente si factoriali in expresiile:}\\\\ 2n+2~si~ 2n-1 ~~\text{care apartin lui N chiar daca } n = 2,5 ~~sau ~~3,5 ~~etc.[/tex]