<s>[INST] Find the value of the constant c such that $\sum_{n=2}^\infty(1+c)^{-n} = 2$ 
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 ...
$$\begin{align}
2 &= \frac{(1+c)^{-2}}{1-(1+c)} \\
2 &= \frac{(1+c)^{-2}}{-c} \\
-2c &= (1+c)^{-2} \\
-2c &= \frac{1}{(1+c)^2} \\
-2c(1+c)^2 &= 1 \\
-2c(1+2c+c^2) &= 1
\end{align}$$
 [/INST] 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, $|1+c|>1$.
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