### Video Transcript

Consider the sequence one, one,
three-quarters, four-eighths, and so on. Which of the following is the
general term of this sequence such that π is greater than or equal to zero? Option (A) π over two to the
power of π. Option (B) π minus one over
two to the power of π. Option (C) π plus one over two
to the power of π. Option (D) two π over two to
the power of π. Or option (E) π plus two over
two to the power of π.

In this question, weβre given a
sequence and asked to find its general term. When weβre finding a general
term, weβre really finding a rule that connects the term number with the actual
value of the term. When weβre given that the index
π is greater than or equal to zero, that means that our sequence begins with
the zeroth term. We then have the first term π
sub one, the second term π sub two, and so on. The πth term would be π sub
π. So given any value of π, what
would the value in the sequence be? If we consider this sequence,
there isnβt a common difference between any two consecutive terms, so this isnβt
an arithmetic sequence. There also isnβt a common ratio
between any two consecutive terms, so this isnβt a geometric sequence
either.

In order to find the general
term of the sequence, weβll have to apply some logic. Letβs take a closer look at
this term, π sub one with the value of one. What if instead of being this
value of one, this value of π sub one was actually a fraction which simplified
to one? In order for a fraction to
simplify to one, the numerator and denominator would have to have the same
value. Letβs say this fraction was
actually two over something, and to simplify to one, it would need to be two
over two. If we think of the zeroth term
π sub zero as instead of just being one as being a fraction of one over one,
now we can see that the numerators actually have quite a nice pattern. They go from one to two to
three to four. The denominators also have a
different pattern. They go from one to two to four
to eight.

Letβs consider the general term
of the numerators and denominators separately for each value of π starting with
π equals zero. Remember that we picked zero
because this was given to us in the question. So for any index π, what will
the numerator be? Well, every value in the
numerator is one more than its index. So the πth term of the
numerator will be π plus one. For the denominators, these
have a pattern which appears to be doubling. In fact, each denominator is a
power of two. It would be two to the power of
π. For example, for the zeroth
term, two to the power of zero gives us one. For the first term, two to the
power of one gives us two and so on. We can now put together the
general term for the numerator and denominator. So the general term of this
sequence is π plus one over two to the power of π, which was the value given
to us in option (C).