### Video Transcript

Multiply the series one divided by
one plus π₯ is equal to the sum from π equals zero to β of negative one to the πth
power π₯ to the πth power by itself to construct a series for one divided by one
plus π₯ squared. Write the answer in sigma
notation.

We recall if we have two power
series, π of π₯ is equal to the sum over π of π π π₯ to the πth power and π of
π₯ is equal to the sum over π of π π π₯ to the πth power, which both converge
for this value of π₯. Then we can calculate the product
of these two power series, π of π₯ multiplied by π of π₯, as the sum over π of
all of our π₯ to the πth power terms with the coefficient the sum over π π π
multiplied by π π minus π. In our case, we want to multiply
one divided by one plus π₯ by itself to get the power series for one divided by one
plus π₯ squared. So we set both of our functions, π
of π₯ and π of π₯, to be equal to one divided by one plus π₯.

So we have both π of π₯ and π of
π₯ are equal to the sum of one over one plus π₯ which are equal to the sum from π
equals zero to β of negative one to the πth power π₯ to the πth power. So for the value of π₯ where our
power series for π of π₯ converges, we can use our formula to find an expression
for π of π₯ squared. The question tells us that one
divided by one plus π₯ is equal to the sum from π equals zero to β of negative one
to the πth power π₯ to the πth power. This gives us that π π is
negative one raised to the power of π. And π π minus π is negative one
raised to the power of π minus π.

Therefore, we can calculate π π
multiplied by π π minus π as negative one to the power of π multiplied by
negative one to the power of π minus π, which is just negative one to the πth
power. So we now have our coefficient of
π₯ to the power of π is the sum from π equals zero to π of negative one to the
πth power. And we can see thatβs our summand,
negative one to the πth power, is independent of π. So, in fact, our sum is negative
one to the πth power added to itself π plus one times. This is because π goes from zero
to π. And if weβre adding negative one to
the πth power to itself π plus one times, this is the same as saying π plus one
multiplied by negative one to the πth power.

Therefore, weβve shown for the
values of π₯ where a power series for one divided by one plus π₯ converges. We can multiply the power series of
one divided by one plus π₯ by itself to get that one divided by one plus π₯ squared
is equal to the sum from π equals zero to β of negative one to the πth power
multiplied by π plus one multiplied by π₯ to the πth power.