Find the Sum of the Series -8+(-4)+(-2)+(-1)+(-1/2)

$- 8 + (- 4) + (- 2) + (- 1) + (- \frac{1}{2})$

Remove parentheses.

$- 8, - 4, - 2, - 1, - \frac{1}{2}$

Remove parentheses.

$- 8, - 4, - 2, - 1, - \frac{1}{2}$

Remove parentheses.

$- 8, - 4, - 2, - 1, - \frac{1}{2}$

Remove parentheses.

$- 8, - 4, - 2, - 1, - \frac{1}{2}$

This is a geometric sequence since there is a common ratio between each term. In this case, multiplying the previous term in the sequence by $\frac{1}{2}$ gives the next term. In other words, $a_{n} = a_{1} \cdot r^{n - 1}$ .

Geometric Sequence: $r = \frac{1}{2}$

This is the form of a geometric sequence.

$a_{n} = a_{1} r^{n - 1}$

Substitute in the values of $a_{1} = - 8$ and $r = \frac{1}{2}$ .

$a_{n} = (- 8) \cdot {(\frac{1}{2})}^{n - 1}$

Simplify the expression.

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Apply the product rule to $\frac{1}{2}$ .

$a_{n} = - 8 \cdot \frac{1^{n - 1}}{2^{n - 1}}$

One to any power is one.

$a_{n} = - 8 \cdot \frac{1}{2^{n - 1}}$

Combine $- 8$ and $\frac{1}{2^{n - 1}}$ .

$a_{n} = \frac{- 8}{2^{n - 1}}$

Move the negative in front of the fraction.

$a_{n} = - \frac{8}{2^{n - 1}}$

This is the formula to find the sum of the first $n$ terms of the geometric sequence. To evaluate it, find the values of $r$ and $a_{1}$ .

$S_{n} = \frac{a_{1} (r^{n} - 1)}{r - 1}$

Replace the variables with the known values to find $S_{0}$ .

$S_{0} = - 8 \cdot \frac{{(\frac{1}{2})}^{0} - 1}{\frac{1}{2} - 1}$

Simplify the numerator.

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Apply the product rule to $\frac{1}{2}$ .

$S_{0} = - 8 \cdot \frac{\frac{1^{0}}{2^{0}} - 1}{\frac{1}{2} - 1}$

Anything raised to $0$ is $1$ .

$S_{0} = - 8 \cdot \frac{\frac{1}{2^{0}} - 1}{\frac{1}{2} - 1}$

Anything raised to $0$ is $1$ .

$S_{0} = - 8 \cdot \frac{\frac{1}{1} - 1}{\frac{1}{2} - 1}$

Divide $1$ by $1$ .

$S_{0} = - 8 \cdot \frac{1 - 1}{\frac{1}{2} - 1}$

Subtract $1$ from $1$ .

$S_{0} = - 8 \cdot \frac{0}{\frac{1}{2} - 1}$

Simplify the denominator.

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To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$S_{0} = - 8 \cdot \frac{0}{\frac{1}{2} - 1 \cdot \frac{2}{2}}$

Combine $- 1$ and $\frac{2}{2}$ .

$S_{0} = - 8 \cdot \frac{0}{\frac{1}{2} + \frac{- 1 \cdot 2}{2}}$

Combine the numerators over the common denominator.

$S_{0} = - 8 \cdot \frac{0}{\frac{1 - 1 \cdot 2}{2}}$

Simplify the numerator.

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Multiply $- 1$ by $2$ .

$S_{0} = - 8 \cdot \frac{0}{\frac{1 - 2}{2}}$

Subtract $2$ from $1$ .

$S_{0} = - 8 \cdot \frac{0}{\frac{- 1}{2}}$

Move the negative in front of the fraction.

$S_{0} = - 8 \cdot \frac{0}{- \frac{1}{2}}$

Multiply the numerator by the reciprocal of the denominator.

$S_{0} = - 8 \cdot (0 (- 1 \cdot 2))$

Multiply $- 1$ by $2$ .

$S_{0} = - 8 \cdot (0 \cdot - 2)$

Multiply by zero.

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Multiply $0$ by $- 2$ .

$S_{0} = - 8 \cdot 0$

Multiply $- 8$ by $0$ .

$S_{0} = 0$

Find the Sum of the Series -8+(-4)+(-2)+(-1)+(-1/2)

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