# What is the solution to this equation: (2n+1)!/(2n-1)! =42 ;

*print*Print*list*Cite

### 1 Answer

You want to solve,

`((2n+1)!)/((2n-1)!) = 42`

I can expand (2n+1)! as below.

`((2n+1)xx2nxx(2n-1)!)/((2n-1)!) = 42`

This gives,

`(2n+1)xx2n = 42`

`(2n+1) xx n = 21`

`2n^2+n-21 = 0`

The solutions for above equation are,

`n = (-1+-sqrt(1+4xx2xx21))/(2xx2)`

`n = (-1+-sqrt(169))/4`

`n = (-1+-13)/4`

`n = 12/4 or n = -14/4`

We know `n!= -14/4` since factorial is not defined for negative numbers.

Therefore,

`n = 12/4 = 3`

**The answer is 3.**