Question

The following is a general method for finding the expresion for the Fibonacci numbers well number of other similar problems. Let (XN) be a sequence of numbers which are defined by the reusion relation XypXn-1+qXN-2. X, X, arbitrary then clear that once Xo and X are given, one can calculate using this relation the values of X. X.... recursively (hence the name recursion relation). Here P and are given numbers For the Fibonacci sequence, p =q=1 and Xo =0,X; = 1 then I must satisfy the (A) Prove that if we take a trial solution of the form XN= equation = px + (b) Let a, b be the two roots of this quadratic equation Show that for any constants AB the sequence XN = A + BEN satisfies the recursion relation. Choose the constants so that X and X, have values (c) Deduce from the above the formula for the general term of the Fibonacci discussed in the text. For the Fibonacci sequence with Xo = 1, X = 3, show that N+ (d) Find a formula for Xn where Xo = 0,X; = 1 XN = 3.XN-1 + XN-2,

Answer 1

XN = p Xwritq XNz Xox, arbitrary XN a) Put XN=aN in equation we get aN - PN-1 + q 2 N-2 2N-2 (2²) = (px + 4) 2-2 since x 80 so x² = para 6) Let a, b be the root of x = px + q 80 x = a or x=b. a XN = 6N so clearly XN = l Since XN= x so clearly XN Combe any linear combination of it XN = A aN + BONO .Xo = A + B X = Aa + Bb so axo = A + Ba and bxo = Ab + Bb. x = A + Bb a Xo-X, = Bla-b) 30 bxo-X, = A (b-a) 80 B= axo - A = bxo-X a-b

Page : Date X X2 = 4, = (0) 7, Xo = 1, x=3 P=q=1 22= +1 x²-x-1=0 - 80 XN= Xo = x = = 1 A (1 + SS N + (1 - )" A + B A (ItsS) + B(1-1) = 3 A+B + (A-B) = 3 į + ICA-B) =3 (-1)= A+ 13 = 1 80 20 = 1 +5 A = 1tr B = VSH op Xne (1735) XN = 3 XN-1 + XN-2 + 1-3)*** Xo = 0, 0,21 d) x² – 3x-1=0 x= 3 1 9 +4

Page Date: x = 3 + J13 XN = A ( 3 + 373 )N+ B (3 - 56 ) N - A+B Le = 0 x = A ( 3 +50) + B (3-513) = 1 A ( 3 +11 ) -A (3-518 ) = 1 ANTB = 1 XN = I (3+518 JN - I (3-513 )