Question
If x = -1, then the value of $$\frac{1}{{{x^{99}}}}$$ + $$\frac{1}{{{x^{98}}}}$$ + $$\frac{1}{{{x^{97}}}}$$ + $$\frac{1}{{{x^{96}}}}$$ + $$\frac{1}{{{x^{95}}}}$$ + $$\frac{1}{{{x^{94}}}}$$ + $$\frac{1}{x}$$ - 1 is?
Answer: Option C
$$\frac{1}{{{x^{99}}}}{\text{ + }}\frac{1}{{{x^{98}}}} + \frac{1}{{{x^{97}}}} + \frac{1}{{{x^{96}}}} + \frac{1}{{{x^{95}}}} + \frac{1}{{{x^{94}}}} + \frac{1}{x} - 1$$
$$ = \frac{1}{{{{\left( { - 1} \right)}^{99}}}}{\text{ + }}\frac{1}{{{{\left( { - 1} \right)}^{98}}}} + \frac{1}{{{{\left( { - 1} \right)}^{97}}}} + \frac{1}{{{{\left( { - 1} \right)}^{96}}}} + $$ $$\frac{1}{{{{\left( { - 1} \right)}^{95}}}} + $$ $$\frac{1}{{{{\left( { - 1} \right)}^{94}}}} + $$ $$\frac{1}{{\left( { - 1} \right)}} - $$ $$1$$
$$\eqalign{
& = - 1 + 1 - 1 + 1 - 1 + 1 + \frac{1}{{ - 1}} - 1 \cr
& = - 2 \cr} $$
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$$\frac{1}{{{x^{99}}}}{\text{ + }}\frac{1}{{{x^{98}}}} + \frac{1}{{{x^{97}}}} + \frac{1}{{{x^{96}}}} + \frac{1}{{{x^{95}}}} + \frac{1}{{{x^{94}}}} + \frac{1}{x} - 1$$
$$ = \frac{1}{{{{\left( { - 1} \right)}^{99}}}}{\text{ + }}\frac{1}{{{{\left( { - 1} \right)}^{98}}}} + \frac{1}{{{{\left( { - 1} \right)}^{97}}}} + \frac{1}{{{{\left( { - 1} \right)}^{96}}}} + $$ $$\frac{1}{{{{\left( { - 1} \right)}^{95}}}} + $$ $$\frac{1}{{{{\left( { - 1} \right)}^{94}}}} + $$ $$\frac{1}{{\left( { - 1} \right)}} - $$ $$1$$
$$\eqalign{
& = - 1 + 1 - 1 + 1 - 1 + 1 + \frac{1}{{ - 1}} - 1 \cr
& = - 2 \cr} $$
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