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Question Number 190061 by sonukgindia last updated on 26/Mar/23

Answered by a.lgnaoui last updated on 29/Mar/23

$$2(n^{n+1}-n!)=5n^2-3n-2$$$$\frac{2\left(n^{n+1}\right)}{n!}-1=\frac{a}{n!}(a=)$$$$\frac{2n\times n^n}{n!}=\frac{5{(n-\frac{3}{2})}^2-\frac{17}{4}}{n!}+1$$$$\frac{2[(n-\frac{3}{2})+\frac{3}{2}]n^n}{n!}=\frac{5{(n-\frac{3}{2})}^2}{n!}-\left(\frac{13}{4n!}\right)$$$$..............$$$$\left(asuivre\right)..$$

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