Problem

Find the value of c such that $ \displaystyle \…

03:21

Problem 75 Hard Difficulty

Find the value of $ c $ if
$ \displaystyle \sum_{n = 2}^{\infty} (1 + c)^{-n} = 2 $

Answer

$c=\frac{-1-\sqrt{3}}{2}$

Related Courses

Calculus 2 / BC

Calculus: Early Transcendentals

Chapter 11

Infinite Sequences and Series

Section 2

Series

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Video Transcript

Let's find the value of C that makes the Siri's equal to so this Siri's here three. Right? This because it's actually geometric weaken. Right, This is one over one plus C to the end cheer measure, and then our equals one over one plus C. And we should eventually make sure that the absolute value of our which is absolute value one over one plus e. But this is less than one. So we have one over one plus c, an absolute value less than one. That means one plus C in the absolute value is bigger than one. And this tells us that we want one placide data that one or one plus see less than minus one. So we either want seed to be bigger than zero. We're over here, see less than negative, too. So we might get more than one answer. But it has to satisfy one of these, too. So let's keep that in mind. See, Has to be bigger than zero. And the reason we want to keep this in mind is in case we have more than one solution, or if even if we have one solution, if it doesn't satisfy one of these, too. For example, C equals negative. One would not be a good choice. It doesn't satisfy one of these, too. So now, assuming C is bigger than zero or seamless the negative two. So this is just to ensure that that flew value bars less than one. We know that the geometric series we have a formula for this. It's the first term of the series which corresponds to any equals two and then we divide by one minus. R. Now let's go to the next page. So the first term when you plug in and equals two and then one minus R, it's got its simplify this what? And then this becomes one over C C plus one and recall that this entire sum was equal to two from the previous page so we can rewrite this equation and then solve this for sea. So we have C equals negative one plus or minus. So this is the quadratic formula. Unless you factor that somehow. So here, B squared, minus four a c. So one, then being plus four times that's going to be a half there and then over to a so it's giving us negative one plus or minus room three over, too. So recall that the radical three is about one point seven. So we're getting approximately negative one plus or minus one point seven over, too. So that's about two point seven over, too, or oops, sorry and see the negative one there. Point seven over, too. Or if we have the minus sign and that's about it. Earth equal to so here recalled. The value of our C had to be either bigger than zero or C was less the negative, too. This value does not work, so we just want to choose the one that we got from the plus side. So here we should on Lee take oh C two b negative one plus the square root of three over, too. Because that one does work, it's point three five, and that one does satisfy one of these, too. So negative one plus room three. All over, too. That's our values for C. That's the on ly one, and that's the final answer

J H.

Related Courses

Calculus 2 / BC

Calculus: Early Transcendentals

Chapter 11

Infinite Sequences and Series

Section 2

Series

Problem

Find the value of c such that $ \displaystyle \…

Problem 75 Hard Difficulty

Find the value of $ c $ if
$ \displaystyle \sum_{n = 2}^{\infty} (1 + c)^{-n} = 2 $

Answer

$c=\frac{-1-\sqrt{3}}{2}$

Related Courses

Calculus: Early Transcendentals

Related Topics

Discussion

Top Calculus 2 / BC Educators

Grace H.

Catherine R.

Samuel H.

Michael J.

Calculus 2 / BC Courses

Integration Review - Intro

Definite vs Indefinite

Recommended Videos

Watch More Solved Questions in Chapter 11

Video Transcript

Calculus: Early Transcendentals

Grace H.

Catherine R.

Samuel H.

Michael J.

Integration Review - Intro

Definite vs Indefinite