Question Number 190061 by sonukgindia last updated on 26/Mar/23
Answered by a.lgnaoui last updated on 29/Mar/23
![2(n^(n+1) −n!)=5n^2 −3n−2 ((2(n^(n+1) ))/(n!))−1=(a/(n!))(a=) ((2n×n^n )/(n!))=((5(n−(3/2))^2 −((17)/4))/(n!))+1 ((2[(n−(3/2))+(3/2)]n^n )/(n!))=((5(n−(3/2))^2 )/(n!))−(((13)/(4n!))) .............. (a suivre)..](https://www.tinkutara.com/question/Q190159.png)
$$\mathrm{2}\left({n}^{{n}+\mathrm{1}} −{n}!\right)=\mathrm{5}{n}^{\mathrm{2}} −\mathrm{3}{n}−\mathrm{2} \\ $$$$\frac{\mathrm{2}\left({n}^{{n}+\mathrm{1}} \right)}{{n}!}−\mathrm{1}=\frac{{a}}{{n}!}\left({a}=\right) \\ $$$$\frac{\mathrm{2}{n}×{n}^{{n}} }{{n}!}=\frac{\mathrm{5}\left({n}−\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\mathrm{17}}{\mathrm{4}}}{{n}!}+\mathrm{1} \\ $$$$\frac{\mathrm{2}\left[\left({n}−\frac{\mathrm{3}}{\mathrm{2}}\right)+\frac{\mathrm{3}}{\mathrm{2}}\right]{n}^{{n}} }{{n}!}=\frac{\mathrm{5}\left({n}−\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} }{{n}!}−\left(\frac{\mathrm{13}}{\mathrm{4}{n}!}\right) \\ $$$$………….. \\ $$$$\left({a}\:{suivre}\right).. \\ $$