### Video Transcript

Find the first five terms of the
sequence π sub π, given π sub π plus one is equal to negative one to the πth
power divided by nine multiplied by π sub π, where π is greater than or equal to
one and π sub one equals negative 11.

In this question, we are given the
value of the first term of our sequence π sub one, which is equal to negative
11. We are asked to find the first five
terms. So we therefore need to calculate
the second, third, fourth, and fifth terms of the sequence. These are denoted π sub two, π
sub three, π sub four, and π sub five.

The formula we are given is an
example of a recursive formula, where π sub π plus one is given in terms of π sub
π. We can therefore calculate any term
of the sequence by substituting in the previous term. The second term π sub two is
therefore equal to negative one to the power of one divided by nine multiplied by π
sub one. We know that π sub one is equal to
negative 11. This expression is equal to
negative one over negative 99, which in turn simplifies to one over 99. This is the second term of the
sequence, and we can use this to calculate the third.

Using our recursive formula once
again, π sub three is equal to negative one squared over nine multiplied by π sub
two. We know that π sub two is one over
99. On the denominator, we have nine
multiplied by one over 99, which is the same as nine over 99. Dividing the numerator and
denominator by nine, this simplifies to one over 11 or one eleventh. As negative one squared is equal to
one, π sub three is equal to one divided by one eleventh. We know that dividing by a fraction
is the same as multiplying by the reciprocal of this fraction. One divided by one eleventh is the
same as one multiplied by 11 over one. This is equal to 11. π sub three, the third term in our
sequence, is 11.

Next, we need to calculate the
fourth term π sub four. This is equal to negative one cubed
divided by nine multiplied by π sub three. And we have just calculated that π
sub three equals 11. The fourth term of the sequence is
therefore equal to negative one over 99, recalling that cubing a negative number
gives a negative answer. Finally, we have π sub five is
equal to negative one to the fourth power divided by nine multiplied by π sub
four. We can substitute negative one over
99 for π sub four. And this in turn simplifies to one
over negative one eleventh. Using a similar method that we used
to calculate π sub three, we have π sub five is equal to negative 11. The first five terms of the
sequence are negative 11, one over 99, 11, negative one over 99, and negative
11.

We notice that the first term is
equal to the fifth term. This means that we have a periodic
sequence where the first four terms are repeated. π sub one is equal to π sub five,
which is equal to π sub nine and so on. Likewise, the second, sixth, and
10th terms are equal. The same is true for the third,
seventh, and 11th terms, together with the fourth, eighth, and 12th terms.