IT has general solution of the form f(n)=r^n
Substituting gives
r^2=2r-1
r^2-2r+1=0
r=1,repeated roots
f(n)=1^n(A+Bn)=A+Bn
f(1)=A+B=1
f(2)=A+2B=5
B=4,A=-3
f(n)=-3+4n
Proof using structural induction
Base case:n=2
f(2)=-3+4*2=5
So base case is true
Inductive hypothesis: Assume true for all 2<=k<=n for some n>=2
Inductive step: We show it is true for n+1
f(n+1)=2f(n)-f(n-1)=2(-3+4n)-(-3+4(n-1))=-6+8n+3-4(n-1)=-3+4n+4=-3+4(n+1)
HEnce true for n+1 and hence for all n>=2
(14) Given a sequence of integers {fi,f2 /. defined by the following recursive function f (n)-.,...
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