If $\frac{(n - 1)! + n!}{(n + 1)!} = \frac{1}{6}$ ${(n-1)!+n!}/((n+1)!) =1/6$ , find $n$ $n$ ?

Question

1861 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-10-10T06:45:16

$n = 6$ $n=6$

$\frac{(n - 1)! + n!}{(n + 1)!} = \frac{1}{6}$ ${(n-1)!+n!}/((n+1)!) =1/6$

$\Leftrightarrow \frac{(n - 1)! + n (n - 1)!}{n (n + 1) \times (n - 1)!} = \frac{1}{6}$ $hArr{(n-1)!+n(n-1)!}/(n(n+1)xx(n-1)!) =1/6$

or $\frac{(n - 1)! (1 + n)}{(n - 1)! (n^{2} + n)} = \frac{1}{6}$ $((n-1)!(1+n))/((n-1)!(n^2+n)) =1/6$

or $\frac{1 + n}{n^{2} + n} = \frac{1}{6}$ $(1+n)/(n^2+n)=1/6$

or $n^{2} + n = 6 + 6 n$ $n^2+n=6+6n$

or $n^{2} - 5 n - 6 = 0$ $n^2-5n-6=0$

i.e. $(n - 6) (n + 1) = 0$ $(n-6)(n+1)=0$

i.e. $n = 6$ $n=6$ or $- 1$ $-1$

But as $(- 1)!$ $(-1)!$ is not defined $n = 6$ $n=6$

If {(n-1)!+n!}/((n+1)!) =1/6(n−1)!+n!(n+1)!=16{(n-1)!+n!}/((n+1)!) =1/6, find nnn?