Question

$\frac{n!+(n+1)!}{(n-1)!+n!}= \frac{48}{7}$

Answer 1

[tex] \frac{(n-1)!n + (n-1)!n(n+1)}{(n-1)! + (n-1)!n} = \frac{48}{7} \\ \\ \frac{(n-1)![n +n (n+1)]}{(n-1)![ 1+ n]} = \frac{48}{7} \\ \\ \frac{n + n(n+1)}{1 +n}= \frac{48}{7} \\ \\ \frac{ n^{2} +2n}{1+n} = \frac{48}{7} \\ \\ 7 n^{2} - 34n - 48=0 \\ \\ n_{1,2} = 6 ; \frac{-8}{7} [/tex]