If n is a positive integer and `U_(n) = (3 + sqrt5)^(n) + (3 - sqrt5)^(n)`, then prove that
`U_(n + 1) = 6U_(n) - 4U_(n -1), n ge 2`

Explanation

L.H.S = `U_(n+1)`
`= (3 + sqrt5)^(n + 1) + (3 - sqrt5)^(n+1)`
`= (3 +sqrt5)^(n) (3 + sqrt5) + (3 - sqrt5)^(n) (3 - sqrt5)`
`= 3 (3 + sqrt5)^(n) + sqrt5 (3 + sqrt5)^(n) + 3 (3 - sqrt5)^(n) - sqrt5 (3 - sqrt5)^(n)`
`= 3 (3 + sqrt5)^(n) + 3 (3 - sqrt5)^(n) + sqrt5 (3 + sqrt5)^(n) - sqrt5 (3 - sqrt5)^(n)`
`= 3 [(3 + sqrt5)^(n) + (3 -sqrt5)^(n)] + sqrt5 [(3 + sqrt5)^(n) - (3 - sqrt5)^(n)]`
L.H.S `= 6U_(n) - 4U_(n -1)`
`= 6 [(3 + sqrt5)^(n) + (3 - sqrt5)^(n)] - 4 [(3 + sqrt5)^(n-1) + (3-sqrt5)^(n-1)]`
`=6 [(3 + sqrt5)^(n) + (3 - sqrt5)^(n)] -4 [((3 + sqrt5)^(n))/((3 + sqrt5)) + ((3 - sqrt5)^(n))/((3 - sqrt5))]`
`=6 [(3 + sqrt5)^(n) + (3 - sqrt5)^(n)]-4 [((3 - sqrt5) (3 + sqrt5)^(n) + (3 + sqrt5) (3 - sqrt5)^(n))/((3 + sqrt5) (3 - sqrt5))]`
`=6 [(3 + sqrt5)^(n) + (3 - sqrt5)^(n)] -4 [(3 (3 + sqrt5)^(n) - sqrt5 (3 + sqrt5)^(n) + 3 (3 - sqrt5)^(n) + sqrt5 (3 - sqrt5)^(n))/((3)^(2) - (sqrt5)^(2))]`
`= 6 [(3 + sqrt5)^(n) + 6(3 - sqrt5)^(n)] -4 [(3(3 + sqrt5)^(n) - sqrt5 (3 + sqrt5)^(n) + 3 (3 - sqrt5)^(n) + 3 (3 -sqrt5)^(n) + sqrt5(3 - sqrt5)^(n))/(4)]`
`= 6 (3 + sqrt5)^(n) + 6 (3 - sqrt5)^(n) -3 (3 + sqrt5)^(n) + sqrt5(3 + sqrt5)^(n) -3 (3 - sqrt5)^(n) - sqrt5 (3 - sqrt5)^(n)`
`= 3 (3 + sqrt5)^(n) + 3 (3 - sqrt5)^(n) + sqrt5 [(3 + sqrt5)^(n) - (3 - sqrt5)^(n)]`
`=3 [(3 + sqrt5)^(n) + (3 - sqrt5)^(n)] + sqrt5 [(3 + sqrt5)^(n) - (3 - sqrt5)^(n)]`
`:.` LHS = RHS
Hence `U_(n+1) = 6U_(n) - 4U_(n-1)`