Question

​ f(1)=−6 f(2)=−4 f(n)=f(n−2)+f(n−1) ​ f(n)=?

Answer 1

By definition, we have

$f(n) = f(n - 1) + f(n - 2)$

so that by substitution,

$f(n-1) = f(n-2) + f(n-3) \implies f(n) = 2f(n-2) + f(n-3)$

$f(n-2) = f(n-3) + f(n-4) \implies f(n) = 3f(n-3) + 2f(n-4)$

$f(n-3) = f(n-4) + f(n-5) \implies f(n) = 5f(n-4) + 3f(n-5)$

$f(n-4) = f(n-5) + f(n-6) \implies f(n) = 8f(n-5) + 5f(n-6)$

and so on.

Recall the Fibonacci sequence $F(n)$, whose first several terms for $n\ge1$ are

$\{1, 1, 2, 3, 5, 8, 13, 21, 34, 55, \ldots\}$

Let $F_n$ denote the $n$-th Fibonacci number. Notice that the coefficients in each successive equation form at least a part of this sequence.

$f(n) = f(n-1) + f(n-2) = F_2f(n-1) + F_1 f(n-2)$

$f(n) = 2f(n-2) + f(n-3) = F_3 f(n-2) + F_2 f(n-3)$

$f(n) = 3f(n-3) + 2f(n-4) = F_4 f(n-3) + F_3 f(n-4)$

$f(n) = 5f(n-4) + 3f(n-5) = F_5 f(n-4) + F_4 f(n-5)$

$f(n) = 8f(n-5) + 5f(n-6) = F_6 f(n-5) + F_5 f(n-6)$

and so on. After $k$ iterations of substituting, we would end up with

$f(n) = F_{k+1} f(n - k) + F_k f(n - (k+1))$

so that after $k=n-2$ iterations,

$f(n) = F_{(n-2)+1} f(n - (n-2)) + F_{n-2} f(n - ((n-2)+1)) \\\\ f(n) = f(2) F_{n-1} + f(1) F_{n-2} \\\\ \boxed{f(n) = -4 F_{n-1} - 6 F_{n-2}}$

Answer 2

The nth term of the sequence is 2n - 8

Equation of a function

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

Learn more on nth term of an AP here: brainly.com/question/19296260

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