Question

Prove, using mathematical induction, that the given property of the Fibonacci numbers is true [F(n+1)]2 =[F(n)]2+F(n−1)∗F(n+2) fora????????n≥2 Where Fibonacci numbers are defined by the recurrence relation F(1) = 1 F(2) = 1 F(n) = F(n − 1) + F(n − 2)

Answer 1

Answer:

We have proven that the property of the Fibonacci sequence $[F(n+1)]^2 =[F(n)]^2+F(n-1)F(n+2)$ holds by Mathematical Induction.

Explanation:

For n≥2. We prove that it holds for n=2, Assume that it holds for n=k and prove that it holds for n=k+1

F(1)=1

F(2) = 1

F(3) = F(3 − 1) + F(3 − 2)= F(2) + F(1)=1+1=2

F(4) = F(4 − 1) + F(4 − 2)= F(3) + F(2)=2+1=3

In $[F(n+1)]^2 =[F(n)]^2+F(n-1)F(n+2)$

We prove it is true for n=2.

When n=2

$L.H.S: [F(n+1)]^2 = [F(2+1)]^2 = [F(3)]^2 = 2^2 =4\\R.H.S: [F(n)]^2+F(n-1)F(n+2) \\=[F(2)]^2+F(2-1)F(2+2)\\= [F(2)]^2+F(1)F(4)\\=1^2+1*3=1+3=4$

We assume it is true for n=k and prove that it holds for n=k+1.

When n=k+1 in $[F(n+1)]^2 =[F(n)]^2+F(n-1)F(n+2)$

Substituting n=k+1 in the LHS: $[F(n+1)]^2$ and applying: F(k+2)=F(k+1)+F(k)

$LHS: [F(k+2)]^2=[F(k+1)+F(k)]^2\\= [F(k+1)]^2+2F(k+1)F(k)+[F(k)]^2\\=[F(k+1)]^2+F(k)[2F(k+1)+F(k)]\\=[F(k+1)]^2+F(k)[F(k+1)+F(k+1)+F(k)]\\=[F(k+1)]^2+F(k)[F(k+1)+F(k+2)]\\=[F(k+1)]^2+F(k)F(k+3)$

Substituting n=k+1 in the RHS :$[F(n)]^2+F(n-1)F(n+2)$

RHS=$[F(K+1)]^2+F(K+1-1)F(K+1+2)=[F(k+1)]^2+F(k)F(k+3)$

Since the LHS=RHS

Therefore, the property is true.

FOR REFERENCE

When n=k in $[F(n+1)]^2 =[F(n)]^2+F(n-1)F(n+2)$

Substituting n=k in the LHS: $[F(n+1)]^2$ and applying: F(k+1)=F(k)+F(k-1)

$LHS: [F(k+1)]^2=[F(k)+F(k-1)]^2\\= [F(k)]^2+2F(k)F(k-1)+[F(k-1)]^2\\=[F(k)]^2+F(k-1)[2F(k)+F(k-1)]\\=[F(k)]^2+F(k-1)[F(k)+F(k)+F(k-1)]\\=[F(k)]^2+F(k-1)[F(k)+F(k+1)]\\=[F(k)]^2+F(k-1)F(k+2)\\Substituting \: n=k \:in\: th\:e RHS :[F(n)]^2+F(n-1)F(n+2)$

$RHS=[F(K)]^2+F(K-1)F(K+2)$

LHS=RHS