### Video Transcript

Consider the series π of π₯ equals
one over one plus π₯, which equals the sum of negative one to the πth power times
π₯ to the πth power, for values of π between zero and β. Differentiate the given series
expansion of π term by term to find the corresponding series expansion for the
derivative of π. Then use the result of the first
part to evaluate the sum of the series the sum of negative one to the π plus one
times π plus one over three to the πth power for values of π between zero and
β.

Weβve been given the series
expansion of π. So letβs work out the first few
terms. The first term is when π is equal
to zero. Itβs negative one to the power of
zero times π₯ to the power of zero, which is equal to one. The second term is when π is equal
to one. So itβs negative one to the power
of one times π₯ to the power of one, which is negative π₯. We then add negative one squared
times π₯ squared, which is π₯ squared. And we continue in this manner. And we see that π of π₯ is equal
to one minus π₯ plus π₯ squared minus π₯ cubed plus π₯ to the fourth power minus π₯
to the fifth power, and so on.

The question tells us to
differentiate the given series expansion of π that will give us π prime. And of course we can simply do this
term by term. The derivative of one is zero, and
the derivative of negative π₯ is negative one. The derivative of π₯ squared is two
π₯. And then our next term is negative
three π₯ squared.

We continue in this way,
multiplying each term by its exponent and then reducing that exponent by one. Without writing the zero, we see
that we can write π prime of π₯ as negative π₯ to the power of zero β remember,
thatβs just negative one β plus two π₯ to the power of one minus three π₯ squared
plus four π₯ cubed, and so on.

Letβs think about how we can write
this as a sum. We know that we start with a
negative term, and then the sign alternates. To achieve this, we need negative
one to the power of π plus one. This will work for values of π
between zero and β. We then multiply each term by π
plus one. So the first term is when π is
equal to zero. And weβre multiplying that by
one. The second term is when π is equal
to one. And weβre multiplying that by two,
and so on. And this is all times π₯ to the
πth power.

And so we found the series
expansion for the derivative of π. Itβs the sum of negative one to the
power of π plus one times π plus one times π₯ to the πth power for values of π
between zero and β.

Part two says to use the result of
the first part to evaluate the sum of the series the sum of negative one to the
power of π plus one times π plus one over three to the πth power for values of π
between zero and β. Letβs compare this sum to the one
we generated in the first part of this question. If we rewrite this as the sum of
negative one to the power of π plus one times π plus one times a third to the πth
power, then we see itβs of the same form. Just π₯ equals a third.

So what weβre going to do is weβre
going to differentiate our original function one over one plus π₯ and evaluate that
when π₯ is equal to one-third. Letβs write π of π₯ as one plus π₯
to the power of negative one. And then we use the chain rule. When we do, we find its derivative
to be equal to negative one over one plus π₯ squared. The sum of our series will be the
value of the derivative when π₯ is equal to one-third. So thatβs negative one over one
plus one-third squared. Thatβs negative one over 16 over
nine, which is simply negative nine sixteenths.