InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
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` |
|
Answer» 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)` |
|