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The nth term of the sequence is 2n - 8
The nth term of an arithmetic progression is expressed as;
Tn = a + (n - 1)d
where
a is the first term
d is the common difference
n is the number of terms
Given the following parameters
a = f(1)=−6
f(2) = −4
Determine the common difference
d = f(2) - f(1)
d = -4 - (-6)
d = -4 + 6
d = 2
Determine the nth term of the sequence
Tn = -6 + (n -1)(2)
Tn = -6+2n-2
Tn = 2n - 8
Hence the nth term of the sequence is 2n - 8
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By definition, we have
[tex]f(n) = f(n - 1) + f(n - 2)[/tex]
so that by substitution,
[tex]f(n-1) = f(n-2) + f(n-3) \implies f(n) = 2f(n-2) + f(n-3)[/tex]
[tex]f(n-2) = f(n-3) + f(n-4) \implies f(n) = 3f(n-3) + 2f(n-4)[/tex]
[tex]f(n-3) = f(n-4) + f(n-5) \implies f(n) = 5f(n-4) + 3f(n-5)[/tex]
[tex]f(n-4) = f(n-5) + f(n-6) \implies f(n) = 8f(n-5) + 5f(n-6)[/tex]
and so on.
Recall the Fibonacci sequence [tex]F(n)[/tex], whose first several terms for [tex]n\ge1[/tex] are
[tex]\{1, 1, 2, 3, 5, 8, 13, 21, 34, 55, \ldots\}[/tex]
Let [tex]F_n[/tex] denote the [tex]n[/tex]-th Fibonacci number. Notice that the coefficients in each successive equation form at least a part of this sequence.
[tex]f(n) = f(n-1) + f(n-2) = F_2f(n-1) + F_1 f(n-2)[/tex]
[tex]f(n) = 2f(n-2) + f(n-3) = F_3 f(n-2) + F_2 f(n-3)[/tex]
[tex]f(n) = 3f(n-3) + 2f(n-4) = F_4 f(n-3) + F_3 f(n-4)[/tex]
[tex]f(n) = 5f(n-4) + 3f(n-5) = F_5 f(n-4) + F_4 f(n-5)[/tex]
[tex]f(n) = 8f(n-5) + 5f(n-6) = F_6 f(n-5) + F_5 f(n-6)[/tex]
and so on. After [tex]k[/tex] iterations of substituting, we would end up with
[tex]f(n) = F_{k+1} f(n - k) + F_k f(n - (k+1))[/tex]
so that after [tex]k=n-2[/tex] iterations,
[tex]f(n) = F_{(n-2)+1} f(n - (n-2)) + F_{n-2} f(n - ((n-2)+1)) \\\\ f(n) = f(2) F_{n-1} + f(1) F_{n-2} \\\\ \boxed{f(n) = -4 F_{n-1} - 6 F_{n-2}}[/tex]