Question

How do you solve `(n+2) ! = 132 n !`?

Answer 1

Simplify `((n+2)!)/(n!)` and solve the resulting quadratic to find `n = 10`

Explanation:

Divide both sides by `n!` to get:

`132 = ((n+2)!)/(n!) = ((n+2)xx(n+1)xx color(red)(cancel(color(black)(n xx ... xx 1))))/(color(red)(cancel(color(black)(n xx ... xx 1))))`

`=(n+2)(n+1) = n^2+3n+2`

If you know your "times table" then you will recognise `132=12 xx 11`, so you can easily spot `n = 10`, but assuming you didn't spot that...

Subtract `132` from both ends to get:

`n^2+3n-130 = 0`

Then find a pair of factors of `130` with difference `3`. The pair `10, 13` works: `13 xx 10 = 130` and `13 - 10 = 3`.

Hence:

`0 = n^2+3n-130 = (n+13)(n-10)`

So `n = 10` or `n = -13`

Ignore the case `n = -13` as factorials of negative numbers are not normally defined. That leaves us with the case `n = 10`.