Answer
$c=-\frac{-1-\sqrt 3}{2}$
Work Step by Step
$\Sigma^{\infty}_{n=2}(1+c)^{-n}=\Sigma^{\infty}_{n=2}\frac{1}{(1+c)^{n}}=2$
$a=\frac{1}{(1+c)^{2}}$ and $r=\frac{1}{1+c}$
$\frac{a}{1-r} =2$
$\frac{\frac{1}{(1+c)^{2}}}{1-\frac{1}{1+c}}=2$
$\frac{\frac{1}{(1+c)^{2}}}{\frac{1+c}{1+c}-\frac{1}{1+c}}=2$
$\frac{\frac{1}{(1+c)^{2}}}{\frac{c}{1+c}}=2$
$\frac{1}{(1+c)^{2}} \times \frac{1+c}{c}=2$
$\frac{1}{(1+c)c}=2$
$\frac{1}{c^{2}+c}=2$
$2c^{2} +2c-1=0$
$c=\frac{-2±\sqrt (2^{2} -4(2)(-1))}{2(2)}$
$c=\frac{-2±\sqrt (4+8)}{4}$
$c=\frac{-2±\sqrt 12}{4}$
$c=\frac{-2±2\sqrt 3}{4}$
$c=\frac{-1±\sqrt 3}{2}$
For the series to converge, we need $|r| = |1+c| \lt 1$
Therefore
$c=-\frac{-1-\sqrt 3}{2}$
