f(1) = 1 and f(n) = 2 · f(n − 1)

Question

what is the 12th term of this sequence?( whole number greater than 1)

Answer 1

f(1) = 1

f(n) = 2 · f(n − 1) for n>1

Often, especially with computers, we start with the value we want to find [ f(12) } and expand that:

f(12) = 2 * f(11)

f(12) = 2 * 2 * f(10)

f(12) = 2 * 2 * 2 * f(9)

f(12) = 2 * 2 * 2 * 2 * f(8)

f(12) = 2 * 2 * 2 * 2 * 2 * f(7)

f(12) = 2 * 2 * 2 * 2 * 2 * 2 * f(6)

f(12) = 2 * 2 * 2 * 2 * 2 * 2 * 2 * f(5)

f(12) = 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * f(4)

f(12) = 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * f(3)

f(12) = 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * f(2)

f(12) = 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * f(1)

Those are all of the recursive cases, now we use the base case, f(1) = 1

f(12) = 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 1

f(12) = 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2

f(12) = 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 4

f(12) = 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 8

f(12) = 2 * 2 * 2 * 2 * 2 * 2 * 2 * 16

f(12) = 2 * 2 * 2 * 2 * 2 * 2 * 32

f(12) = 2 * 2 * 2 * 2 * 2 * 64

f(12) = 2 * 2 * 2 * 2 * 128

f(12) = 2 * 2 * 2 * 256

f(12) = 2 * 2 * 512

f(12) = 2 * 1024

f(12) = 2048

Answer 2

Greetings! Lets solve this shall we?

So, we must solve this sequence to find the 12th term. Notice that if we let n = 1, we have

f(1) = 2*f(1-1) = 2*f(0)..... We do not have the f(0) term so we must begin with f(2) such that

f(2) = 2*f(2-1) = 2*(1) = 2

f(3) = 2*f(3-1) = 2*(2) = 4

f(4) = 2*f(4-1) = 2*(4) = 8

f(5) = 2*f(5-1) = 2*(8) = 16.........As you can see each term is doubling from the previous. Continue this sequence and you will find out that

f(12) = 2*f(12-1) = 2*f(11) = 2*(1024) = 2048

I hope this helped!

Answer 3

Hi,

f(1) = 1 and f(n) = 2 · f(n − 1)

n = 2 f(2) = 2 *f (1) = 2 * 1 = 2

n = 3 f(3) = 2 * f(2) = 2* 2 = 4

n = 4 f(4) = 2 * f(3) = 2* 4 = 8

and so on ......

n= 12 f(12) = 2048

another method you can find a pattern

1, 2, 4, 8, 16,.........2048.

I hope it is useful.

Minoo

Answer 4

f(n)=2ⁿ-1, therefore f(12)=4095

Recursive equations like this one are so-called difference equations.

The general equation of which this is an example can be written

(E-a)f(n) = 0 and have the general solution f(n) = aⁿ + C.