Question

Question #33988

Answer 1

Direct proof and solve

Explanation:

Let `k=n` then

`("_(n-1)^n) + 1` = `("_n^(n+1))`

then we just solve the equation satisfying `k=n`

`(n(n-1)!)/((n-n-1)!(n-1)!) + 1 = ((n+1)!)/((n+1-n)!(n)!)`
simplify
`n + 1 = n+1`
`n = n`
thus
`n = k`

now we solve the case when `k < n` Let `k=n-1`

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

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

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

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

`n^2+ n= n^2 + n`

`n = n`
thus
`n = k`