Chapter 8.2, Problem 53E

To determine

To find: The value of c.

Answer to Problem 53E

  $\begin{array}{l}c=0.366\\ =-1.366\\ =-2\end{array}\}$

Explanation of Solution

Given information: ${\displaystyle \sum _{n=2}^{\infty }{(1+c)}^{-n}}=2$

${\displaystyle \sum _{n=2}^{\infty }{(1+c)}^{-n}}=2$

  $\begin{array}{l}$ \Rightarrow {(1+c)}^{-2}+{(1+c)}^{-3}+{(1+c)}^{-4}+{(1+c)}^{-5}+\mathrm{......}=2\mathrm{......}\text{(1)}$$ \Rightarrow {(1+c)}^{-2}+{(1+c)}^{-3}+{(1+c)}^{-2}[{(1+c)}^{-2}+{(1+c)}^{-3}+{(1+c)}^{-4}+\mathrm{......}]=2$$ \Rightarrow {(1+c)}^{-2}+{(1+c)}^{-3}+{(1+c)}^{-2}\times 2=2\text{ [From equation (1)]}$$ \Rightarrow {(1+c)}^{-2}[1+{(1+c)}^{-1}+2]=2$$ \Rightarrow {(1+c)}^{-2}[{(1+c)}^{-1}+3]=2$$ \Rightarrow {(1+c)}^{-2}[\frac{1}{1+c}+3]=2$$ \Rightarrow {(1+c)}^{-2}[\frac{1+3c+3}{1+c}]=2$$ \Rightarrow {(1+c)}^{-2}[\frac{4+3c}{1+c}]=2$$ \Rightarrow {(1+c)}^{-3}(4+3c)=2$$ \Rightarrow \frac{4+3c}{{(1+c)}^3}=2$$ \Rightarrow 4+3c=2{(1+c)}^3$\Rightarrow 4+3c=2(1+c^3+3c+3c^2)\\ \Rightarrow 4+3c=2(c^3+3c^2+3c+1)\\ \Rightarrow 4+3c=2c^3+6c^2+6c+2\\ \Rightarrow 2c^3+6c^2+6c-2=0\\ \begin{array}{l}\therefore c=0.366\\ =-1.366\\ =-2\end{array}\}\end{array}$

Hence, $\begin{array}{l}c=0.366\\ =-1.366\\ =-2\end{array}\}$