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$(x^{n} + x^{n-1} + ... + x^{2} + x + \frac{1}{x} + \frac{1}{x^{2}} + ... + \frac{1}{x^{n-1}} + \frac{1}{x^{n}})(x^{n+1} + \frac{1}{x^{n+1}})$

$= x^{2n+1} + x^{2n} + ... + x^{n+2} + x^{n} + ... + x^{2} + x + \frac{1}{x} + \frac{1}{x^{2}} + \frac{1}{x^{n}} + ... + \frac{1}{x^{n+2}} + ... + \frac{1}{x^{2n}} + \frac{1}{x^{2n+1}}$

$a = 2n + 1, b = n+1, c = n$

$\begin{aligned} I &\equiv \sum_{n=1}^{a} S_n \\ &= \sum_{n=1}^{a} \left(x^n+\frac{1}{x^n}\right) \\ &= \sum_{n=1}^{a} x^n+\sum_{n=1}^{a} \left(\frac{1}{x}\right)^n \\ &= \frac{x^{a+1}-x}{x-1}+\frac{(1/x)^{a+1}-(1/x)}{(1/x)-1} \\ &= \frac{x^{a+1}-x}{x-1}+\frac{1-x^{-a}}{x-1} \\ &= \frac{(1-x^{-a})(x^{a+1}+1)}{x-1} \end{aligned}$

Now...

$\begin{aligned} \sum_{n=1}^{c} S_n &\equiv \frac{I-S_b}{S_b} \\ \frac{(1-x^{-c})(x^{c+1}+1)}{x-1} &\equiv \frac{(1-x^{-a})(x^{a+1}+1)}{(x-1)(x^b+x^{-b})}-1 \end{aligned}$

So, moving everything to one side and multiplying by $(x-1)(x^b+x^{-b})$ ...

$\begin{aligned} (1-x^{-a})(x^{a+1}+1)-(x-1)(x^b+x^{-b})-(1-x^{-c})(x^{c+1}+1)(x^b+x^{-b}) &\equiv 0 \\ 1-x-x^{-a}+x^{a+1}+x^{-b-c}+x^{b-c}-x^{1-b+c}-x^{1+b+c} &\equiv 0 \end{aligned}$

As all of our terms have coefficient 1 or -1 and two terms may only be identically opposite if they have the same power, the terms must cancel pairwise. The terms with coefficient 1 are $1$ , $x^{a+1}$ , $x^{-b-c}$ , and $x^{b-c}$ . The terms with coefficient -1 are $-x$ , $-x^{-a}$ , $-x^{-b+c+1}$ , and $-x^{b+c+1}$ . We have that $-b-c < 0$ and $-a < 0$ , while $b+c+1 > 0$ , $a+1 > 0$ , and (of course) $1 > 0$ . We have one term with a positive coefficient and a power of $0$ . We have only one term with a negative coefficient and a power that is not determined to be either greater or less than $0$ : $-x^{-b+c+1}$ . To cancel the term $1$ , we must thus have $-b+c+1 = 0$ , or $c = b-1$ . Thus, $b-c = 1 > 0$ . We have only one term with a positive coefficient and a negative power: $x^{-b-c}$ . We have only one term with a negative coefficient and a negative power: $-x^{-a}$ . Thus, these two terms must cancel each other, and $-b-c = -a$ , or $a-b-c = \boxed{0}$ .

Note that the problem statement does not hold true when $x = 1$ . For $x = 1$ , our identity for the geometric sum is false, and we have $\sum_{n=1}^a S_n = 2a$ , $\sum_{n=1}^c S_n = 2c$ , and $S_b = 2$ , so our equation gives us the solution $c = \frac{a-1}{2}$ , which is not valid for even $a$ . I therefore recommend that you specify that $x \neq 1$ , although I assume that most people could figure that out on their own.