Is $|r| < 1$ ?

Algebra Level 3

$\displaystyle \large \sum_{n \ = \ 2}^{\infty} \left(1 + c\right)^{- n} = 2$

What value of $c$ makes the above equation true? If the answer can be expressed as $\dfrac{\sqrt{p} - q}{r}$ for positive integers $p$ , $q$ , and $r$ , and square-free $p$ , find $(p + 1)(q + 2)(r + 3)$ .

The answer is 60.

1 solution

Chew-Seong Cheong
Jul 29, 2017

$\begin{aligned} \sum_{n=2}^\infty (1+c)^{-n} & = 2 \\ \frac 1{(1+c)^2}\sum_{n=0}^\infty \left(\frac 1{1+c}\right)^n & = 2 \\ \frac 1{(1+c)^2} \left(\frac 1{1-\frac 1{1+c}} \right) & = 2 & \small \color{#3D99F6} \text{where }-1 <\frac 1{1+c} < 1 \\ \frac 1{(1+c)^2} \left(\frac {1+c}c \right) & = 2 \\ \frac 1{c+c^2} & = 2 \\ 2c^2 + 2c - 1 & = 0 \\ c & = \frac {-2 \pm \sqrt{4+8}}4 \\ & = \frac {\sqrt 3-1}2 & \small \color{#3D99F6} \frac {-\sqrt 3-1}2 < -1 \text{ rejected.} \end{aligned}$

$\implies (p+1)(q+2)(r+3) = (3+1)(1+2)(2+3) = \boxed{60}$

For sake of completeness, don't forget to check if the series does, in fact, converge with the value of $c$ :)

Zach Abueg - 3 years, 10 months ago

Or perhaps this is already accounted for when you state $- 1 < \dfrac{1}{1 + c} < 1$ . How did I not realize - sorry about that!

Zach Abueg - 3 years, 10 months ago

Is ∣ r ∣ < 1 |r| < 1 ∣ r ∣ < 1 ?

The answer is 60.

1 solution

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Is $|r| < 1$ ?