we shall prove that using induction

f(n)=(1/8)*(13)^n - (1/8)*(5)^n

its true for n=0 and 1
suppose its true for n=1..k, k>=1
then
18f(n – 1) – 65f(n – 2)
=(9/4)13^(n-1)-(9/4)5^(n-1)-(65/8)(13^(n-2))+(65/8)(5^(n-2))
=13^(n-2)(9*13/4-65/8)-5^(n-2)(45/4-65/8)
=(1/8)13^n-(1/8)5^n
=f(n)
proven by induction f(n)=(1/8)*(13)^n - (1/8)*(5)^n for all integer n>=0

So f(1000)= (1/8)*(13)^1000 - (1/8)*(5)^1000

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