, easy induction proof.
I know one way . . . it's rather long . . .
The first term is: . . . . The second term is: .Is there a way of turning recursive equations of the form .
where into an explicit form with variables ?
. . and we have: .
. . which factors: .
. . and has roots: .
Hence, the function contains: .
The function is of the form: .
We determine and by using the first two terms of the sequence.