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

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Answer 1 · 2017-10-10T06:45:16

$n=6$

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

$hArr{(n-1)!+n(n-1)!}/(n(n+1)xx(n-1)!) =1/6$

or $((n-1)!(1+n))/((n-1)!(n^2+n)) =1/6$

or $(1+n)/(n^2+n)=1/6$

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

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

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

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

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

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