Find the Sum of the Series 1+1/2+1/4+1/8

$1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8}$

Step 1

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

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

Step 2

This is the form of a geometric sequence.

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

Step 3

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

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

Step 4

Multiply ${(\frac{1}{2})}^{n - 1}$ by $1$ .

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

Step 5

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

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

Step 6

One to any power is one.

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

Step 7

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 $a_{1}$ .

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

Step 8

Replace the variables with the known values to find $S_{4}$ .

$S_{4} = 1 \cdot \frac{{(\frac{1}{2})}^{4} - 1}{\frac{1}{2} - 1}$

Step 9

Multiply $\frac{{(\frac{1}{2})}^{4} - 1}{\frac{1}{2} - 1}$ by $1$ .

$S_{4} = \frac{{(\frac{1}{2})}^{4} - 1}{\frac{1}{2} - 1}$

Step 10

Multiply the numerator and denominator of the complex fraction by $2$ .

Multiply $\frac{{(\frac{1}{2})}^{4} - 1}{\frac{1}{2} - 1}$ by $\frac{2}{2}$ .

$S_{4} = \frac{2}{2} \cdot \frac{{(\frac{1}{2})}^{4} - 1}{\frac{1}{2} - 1}$

Combine.

$S_{4} = \frac{2 ({(\frac{1}{2})}^{4} - 1)}{2 (\frac{1}{2} - 1)}$

Step 11

Apply the distributive property.

$S_{4} = \frac{2 {(\frac{1}{2})}^{4} + 2 \cdot - 1}{2 (\frac{1}{2}) + 2 \cdot - 1}$

Step 12

Cancel the common factor of $2$ .

Cancel the common factor.

$S_{4} = \frac{2 {(\frac{1}{2})}^{4} + 2 \cdot - 1}{2 (\frac{1}{2}) + 2 \cdot - 1}$

Rewrite the expression.

$S_{4} = \frac{2 {(\frac{1}{2})}^{4} + 2 \cdot - 1}{1 + 2 \cdot - 1}$

Step 13

Simplify the numerator.

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

$S_{4} = \frac{2 (\frac{1^{4}}{2^{4}}) + 2 \cdot - 1}{1 + 2 \cdot - 1}$

Cancel the common factor of $2$ .

Factor $2$ out of $2^{4}$ .

$S_{4} = \frac{2 (\frac{1^{4}}{2 \cdot 2^{3}}) + 2 \cdot - 1}{1 + 2 \cdot - 1}$

Cancel the common factor.

$S_{4} = \frac{2 (\frac{1^{4}}{2 \cdot 2^{3}}) + 2 \cdot - 1}{1 + 2 \cdot - 1}$

Rewrite the expression.

$S_{4} = \frac{\frac{1^{4}}{2^{3}} + 2 \cdot - 1}{1 + 2 \cdot - 1}$

One to any power is one.

$S_{4} = \frac{\frac{1}{2^{3}} + 2 \cdot - 1}{1 + 2 \cdot - 1}$

Raise $2$ to the power of $3$ .

$S_{4} = \frac{\frac{1}{8} + 2 \cdot - 1}{1 + 2 \cdot - 1}$

Multiply $2$ by $- 1$ .

$S_{4} = \frac{\frac{1}{8} - 2}{1 + 2 \cdot - 1}$

To write $- 2$ as a fraction with a common denominator, multiply by $\frac{8}{8}$ .

$S_{4} = \frac{\frac{1}{8} - 2 \cdot \frac{8}{8}}{1 + 2 \cdot - 1}$

Combine $- 2$ and $\frac{8}{8}$ .

$S_{4} = \frac{\frac{1}{8} + \frac{- 2 \cdot 8}{8}}{1 + 2 \cdot - 1}$

Combine the numerators over the common denominator.

$S_{4} = \frac{\frac{1 - 2 \cdot 8}{8}}{1 + 2 \cdot - 1}$

Simplify the numerator.

Multiply $- 2$ by $8$ .

$S_{4} = \frac{\frac{1 - 16}{8}}{1 + 2 \cdot - 1}$

Subtract $16$ from $1$ .

$S_{4} = \frac{\frac{- 15}{8}}{1 + 2 \cdot - 1}$

Move the negative in front of the fraction.

$S_{4} = \frac{- \frac{15}{8}}{1 + 2 \cdot - 1}$

Step 14

Simplify the denominator.

Multiply $2$ by $- 1$ .

$S_{4} = \frac{- \frac{15}{8}}{1 - 2}$

Subtract $2$ from $1$ .

$S_{4} = \frac{- \frac{15}{8}}{- 1}$

Step 15

Reduce the expression by cancelling the common factors.

Dividing two negative values results in a positive value.

$S_{4} = \frac{\frac{15}{8}}{1}$

Divide $\frac{15}{8}$ by $1$ .

$S_{4} = \frac{15}{8}$

Find the Sum of the Series 1+1/2+1/4+1/8

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