Equation with factorials

Write the left hand side as $\frac{(n-1)!}{(n-1)! (n(n+1)-n)}=\frac{1}{n(n+1)-n}.$ So we need to solve $n(n+1)-n=81.$ Since $n(n+1)-n=n^2$ we see that $n=9$ is the answer we were looking for.

We can solve it like this,

$\frac{(n-1)!}{(n+1)\times n! - n!} = \frac{1}{81}$

$\Rightarrow \frac{(n-1)!}{n!\times (n+1-1)} =\frac{1}{81}$

$\Rightarrow \frac{(n-1)!}{n\times(n-1)! \times n} = \frac{1}{81}$

$\Rightarrow \frac{1}{n\times n} = \frac{1}{81}$

$\Rightarrow \frac{1}{n^{2}} = \frac{1}{81}$

Which means,

$n^{2} = 81$

Taking square root on both sides we get ,

$n = \pm9$

Since $n$ is given to be a positive integer , we get

$n =\boxed {9}$

i fliped the equation upside down both sides first.

((n+1)!-n!)/(n-1)! = 81

(n+1)!/(n-1)! - n!/(n-1)! =81

(n-1)!n(n+1)/(n-1)! - (n-1)!n/(n-1)! =81

n(n+1)-n =81

n^2 +n-n=81

n^2=81=9×9

n=9

(n/n!)/[n!((n+1)-1)]--->1/n^2=1/81 ---->n=9

Cross multiply to get : 81*(n-1)! = (n+1)! - n! now divide all the equation by n! L.H.S :
(n-1)!/n! = 1/n THEREFORE 81/n R.H.S: (n+1)!/n! = n + 1 AND n!/n! = 1 therefore : 81/n = n+1 -1 81/n = n 81 = n^2 n = sqrt (81) = 9

(n-1)!/[(n+1)!-n!]=1/81

(n-1)!/[(n+1).n!-n!]=1/81

(n-1)!/[n!.(n+1-1)]=1/81

(n-1)!/[(n-1)!.n²]=1/81

1/n²=1/81

n²=81 n=9

[(n+1)!-n!]/(n-1)!=81 (n+1)!/(n-1)!-n!/(n-1)!=81 n(n+1)-n=81 n×n=81 n=9

81(n-1)!=(n+1)!-n! And 81(n-1)!=(n+1)(n)(n-1)!-(n)(n-1)! And 81=(n+1)(n)-(n) and 81=n^2 and N=9