Question

Find the 10th term of the sequence defined by the given rule. Assume that the domain of each function is the set of whole numbers greater than 0. f(1) = 40, f(n) = 2 × f(n − 1) − 60

Answer 1

f(1)=40

f(2) = 2(40)-60 = 20

f(3) = 2(20)-60 = -20

f(4) = 2(-20)-60= -100

f(5) = 2(-100) - 60 = -260

f(6) = 2(-260) - 60 = -580

We could clearly go on like this and get to f(10) but I suspect we're supposed to find the general form.

Each step doubles and subtracts  so we expect a form

$f(n) = a 2^n + b$

$f(1)=40 = 2a + b$

$f(2) =20 = 4a + b$

Subtracting,

$-20 = 2a$

$a = -10$

$40 = 2a + b = -20 + b$

$b = 60$

We suspect

$f(n) =60 -10 (2^n)$

Let's check:

$f(3) = 60 - 10(8) = -20 \quad\checkmark$

$f(4) = 60 - 10(16) = -100 \quad\checkmark$

$f(5) = 60 - 10(32) = -260 \quad\checkmark$

It looks like we have the correct formula

$f(10) = 60 - 10(2^{10}) = 60 - 10240 = -10180$

Answer: -10180