<s>[INST] Sum of the series $\sum_{k=2}^\infty (-1)^k (\frac{x+1}{2})^k$ Find the sum of the series for those values $x$ of
$$\sum_{k=2}^\infty (-1)^k \left(\frac{x+1}{2}\right)^k$$
I know that
$$a=\frac{x+1}{2}$$ and $$r=\frac{-x-1}{2}$$
When I plug these values into the formula $$S_n=\frac{a(1-r^n)}{1-r},$$ I get $$S_n=\frac{(x+1)(2^n-(-x-1)^n)}{(2^n)(x+3)}$$ However, I am not sure if I did it correctly. Would this be right?
Correction:
$a=\frac{x^2+2x+1}{4}$
So $Sn=\frac{x^2+2x+1}{2x+6}$ and it is correct when -3 < x < 1
 [/INST] Since it says $\displaystyle\sum_{k=2}^\infty$ rather than $\displaystyle\sum_{k=1}^\infty$, the first term is $a= (-1)^2 \left( \dfrac {x+1} 2 \right)^2$.  The common ratio is exactly what you had, $r=\dfrac{-x-1}2$.  The sum of the infinite geometric series with first term $a$ and common ratio $r$ is given by
$$
a + ar + ar^2 + ar^3 + \cdots = \frac a {1-r}.
$$ 
The expression you gave, $S_n = \dfrac{a(1-r^n)}{1-r}$ is for a finite geometric series with $n$ terms.  You said
$$
a = \text{something}
$$
and
$$
r = \text{something}
$$
but you omitted
$$
n=\text{something},
$$
and that's because in this problem there is no $n$ because the number of terms is infinite.
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