### Video Transcript

Find in terms of π the general
term of the sequence one, two, four, eight, and so on where π is greater than or
equal to one.

In this question, weβre asked to
find the general or πth term of the sequence. So letβs begin by establishing what
type of sequence we have.

If there is a common difference
where the same value is added or subtracted between any two terms, then we would
have an arithmetic sequence. Therefore, letβs see if we can
calculate the difference between the first and second terms. Well, we can get to the second term
from the first term by adding one. However, to go from the second to
the third term, we must add two. And from the third to the fourth
term, we must have added four. Therefore, this is not an
arithmetic sequence.

The next thing we might choose to
do is to establish if we have a geometric sequence. In a geometric sequence, there is a
fixed ratio between successive terms. Letβs calculate the ratio between
the first and second terms. And we can easily see that if we
multiply the first term, one, by two, we would get the second term of two. We could also get from the second
term to the third term by multiplying by two. And finally, we can see that the
ratio between the third term and the fourth term is also two, which means that we do
indeed have a geometric sequence.

We can remember that the πth term
π sub π of a geometric sequence is calculated as π times π to the power of π
minus one, where π or π sub one is the first term and π is the fixed ratio
between terms. In this sequence, we have
calculated that the ratio between terms is two. So that means that π equals
two. The value of π is simply the first
term in the sequence, which is one. When we substitute these values
into the πth term formula, we get π sub π is equal to one times two to the power
of π minus one. We can then simplify this answer to
give the general term of the sequence one, two, four, eight, and so on as to two the
power of π minus one.