Question

Which recursive formula can be used to generate the sequence below, where f(1) = 3 and n ≥ 1?

3, –6, 12, –24, 48, …

f (n + 1) = –3 f(n )
f (n + 1) = 3 f(n )
f (n + 1) = –2 f(n )
f (n + 1) = 2 f(n)

Answer 1

f(1)=3
f(2)=-6
f(3)=12
f(4)=-24

It is very clear from observing the pattern that f(n+1)=-2f(n)
So, option C is correct.

Answer 2

Answer:

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

Step-by-step explanation:

Given : 3, –6, 12, –24, 48, …

f(1)= first term = 3

Common ratio = r = $\frac{-6}{3} =\frac{12}{-6} =-2$

Since the given sequence forms the G.P.

So, formula of nth term in G.P. = $f(n)=f(1) \times r^{n-1}$

So, the formula fro the given sequence of nth term :

$f(n)=3 \times (-2)^{n-1}$

Now,  $f(n+1)=3 \times (-2)^{n+1-1}$

$f(n+1)=3 \times (-2)^{n}$

So, recursive formula:

$\frac{f(n+1)}{f(n)}=\frac{3 \times (-2)^{n}}{3 \times (-2)^{n-1}}$

$\frac{f(n+1)}{f(n)}=(-2)^1$

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

Hence recursive formula can be used to generate the sequence below, where f(1) = 3 and n ≥ 1 is  $f(n+1) = -2\times f(n)$

Thus Option c is correct.