Video Transcript
Evaluate the ∑ from 𝑟 equals eight
to ∞ of thirty-five fourths times negative one-half to the 𝑟 minus one power.
This sigma notation can be used to
represent infinite series. If the sequence goes to infinity,
it continues without end; it doesn’t stop. The infinite symbol is placed above
the summation symbol to indicate that the series is infinite and the number at the
bottom is what we plug in to our expression first. So 𝑟 equals eight and then any
number after eight, we keep plugging in. So we are not plugging in any
number less than eight.
So let’s think for example if it
didn’t go to infinity; let’s say it went to 10. So it goes from 𝑟 equals 8 until
𝑟 equals 10. So we can plug in eight, get a
number, plug in nine, get a number, and plug in 10, get a number and add them
together. And that would be our
summation. But since this goes to infinity, we
can’t just plug in a bunch of numbers. We’re going to have to use a
formula. The formula we would need to use
would be 𝑎 sub one divided by one minus 𝑟. 𝑎 sub one would be the very first
term that we would get when we plug in our first 𝑟, which is 𝑟 equals eight.
So whenever we plug in eight, we
will find what our first term would be, 𝑎 sub one. And then 𝑟 on the denominator is
different than the 𝑟 equals eight; that tells us the very first number to plug
in. This 𝑟 is located here. So let’s begin by first finding 𝑎
sub one. And we’ll do that by plugging in
the very first number we are allowed to; that is, eight. Thirty-five fourths times negative
one-half to the eight minus one power and eight minus one is seven. So we need to take negative
one-half to the seventh power, which means we take negative one to the seventh power
and two to the seventh power, which results in negative one over 128.
And now, we multiply these
fractions. And when we multiply fractions, we
multiply the numerators together, which gives us negative 35, and we multiply the
denominators together, which gives us 512. So we can plug in negative 35 over
512 in for 𝑎 sub one. And we already know what 𝑟 should
be; it’s negative one-half. So we have plugged in 𝑎 sub one
and we have plugged in 𝑟. So on the denominator, one minus
negative one-half means we are really adding one and one-half, which is one and a
half, which to write as an improper fraction, that would be three-halves.
Now, when dividing fractions, we
actually would change this to multiplication and multiply by the denominator’s
reciprocal. So we flip the bottom, the
denominator. So before multiplying straight
across, two can go into 512 256 times. Now, we can multiply straight
across because nothing else simplifies. So the two actually went into
itself once, so negative 35 times one for the numerator, which is negative 35. And our denominator 256 times three
would be 768. So this would be our final
answer.
