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

Question

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

For this question,I'm not sure if I'm doing it right. If I am doing it right, I'm not sure how to get further. Here is what I have so far. Can anyone please help me out?

Comparing the series to the conventional form $a + a r + a r^2 + \cdots = a/(1-r)$, we have

$$r = 1+c \qquad\text{and}\qquad a = (1+c)^{-2} \qquad\text{and}\qquad 2 = \frac{a}{1-r}$$ So ... $$\begin{align} 2 &= \frac{(1+c)^{-2}}{1-(1+c)} \\ 2 &= \frac{(1+c)^{-2}}{-c} \\ -2c &= (1+c)^{-2} \\ -2c &= \frac{1}{(1+c)^2} \\ -2c(1+c)^2 &= 1 \\ -2c(1+2c+c^2) &= 1 \end{align}$$

Answer 1

HINT: you must solve the equation $$\frac{1}{c(1+c)}=2$$ for $c$ this comes from $$\sum_{i=2}^n(1+c)^{-i}=\frac{(c+1)^{-n-1} \left((c+1)^n-c-1\right)}{c}$$

Answer 2

When the series converges,

$$\sum_{n=2}^\infty(1+c)^{-n}=\frac{(1+c)^{-2}}{1-(1+c)^{-1}}=\frac1{(1+c)\,c}=2.$$

Solve $$c^2+c-\frac12=0.$$

After solving, check the convergence condition, $|1+c|>1$.