QuestionCalcula la suma: $\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}=$

Studdy Solution

STEP 1

Assumptions1. We are working with real numbers. . The operations are basic arithmetic operations (addition and subtraction).
3. The fractions are all in the form of $\frac{1}{^n}$ where $n$ is a positive integer.

STEP 2

We will start by adding and subtracting the fractions as they are. The operation between two fractions can be performed by finding a common denominator and then adding or subtracting the numerators.

STEP 3

The common denominator for $\frac{1}{2}$ and $\frac{1}{}$ is $$. So we convert $\frac{1}{2}$ to $\frac{2}{}$.
$\frac{1}{2} = \frac{2}{}$

STEP 4

Now subtract $\frac{1}{4}$ from $\frac{2}{4}$ .
$\frac{2}{4} - \frac{1}{4} = \frac{1}{4}$

STEP 5

The common denominator for $\frac{1}{4}$ and $\frac{1}{8}$ is $8$ . So we convert $\frac{1}{4}$ to $\frac{2}{8}$ .
$\frac{1}{4} = \frac{2}{8}$

STEP 6

Now add $\frac{1}{8}$ to $\frac{2}{8}$ .
$\frac{2}{8} + \frac{1}{8} = \frac{3}{8}$

STEP 7

The common denominator for $\frac{3}{}$ and $\frac{1}{16}$ is $16$ . So we convert $\frac{3}{}$ to $\frac{6}{16}$ .
$\frac{3}{} = \frac{6}{16}$

STEP 8

Now subtract $\frac{1}{16}$ from $\frac{6}{16}$ .
$\frac{6}{16} - \frac{1}{16} = \frac{5}{16}$ So, the simplified form of the given expression $\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}$ is $\frac{5}{16}$ .

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QuestionCalcula la suma: $\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}=$

Studdy Solution

STEP 1

Assumptions1. We are working with real numbers. . The operations are basic arithmetic operations (addition and subtraction).
3. The fractions are all in the form of $\frac{1}{^n}$ where $n$ is a positive integer.

STEP 2

We will start by adding and subtracting the fractions as they are. The operation between two fractions can be performed by finding a common denominator and then adding or subtracting the numerators.

STEP 3

The common denominator for $\frac{1}{2}$ and $\frac{1}{}$ is $$. So we convert $\frac{1}{2}$ to $\frac{2}{}$.
$\frac{1}{2} = \frac{2}{}$

STEP 4

Now subtract $\frac{1}{4}$ from $\frac{2}{4}$ .
$\frac{2}{4} - \frac{1}{4} = \frac{1}{4}$

STEP 5

The common denominator for $\frac{1}{4}$ and $\frac{1}{8}$ is $8$ . So we convert $\frac{1}{4}$ to $\frac{2}{8}$ .
$\frac{1}{4} = \frac{2}{8}$

STEP 6

Now add $\frac{1}{8}$ to $\frac{2}{8}$ .
$\frac{2}{8} + \frac{1}{8} = \frac{3}{8}$

STEP 7

The common denominator for $\frac{3}{}$ and $\frac{1}{16}$ is $16$ . So we convert $\frac{3}{}$ to $\frac{6}{16}$ .
$\frac{3}{} = \frac{6}{16}$

STEP 8

Now subtract $\frac{1}{16}$ from $\frac{6}{16}$ .
$\frac{6}{16} - \frac{1}{16} = \frac{5}{16}$ So, the simplified form of the given expression $\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}$ is $\frac{5}{16}$ .

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QuestionCalcula la suma: 12−14+18−116=\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}=21​−41​+81​−161​=

Studdy Solution

STEP 1

Assumptions1. We are working with real numbers. . The operations are basic arithmetic operations (addition and subtraction).3. The fractions are all in the form of 1n\frac{1}{^n}n1​ where nnn is a positive integer.

STEP 2

We will start by adding and subtracting the fractions as they are. The operation between two fractions can be performed by finding a common denominator and then adding or subtracting the numerators.

STEP 3

The common denominator for 12\frac{1}{2}21​ and 1\frac{1}{}1​ is $$. So we convert $\frac{1}{2}$ to $\frac{2}{}$.12=2\frac{1}{2} = \frac{2}{}21​=2​

STEP 4

Now subtract 14\frac{1}{4}41​ from 24\frac{2}{4}42​.24−14=14\frac{2}{4} - \frac{1}{4} = \frac{1}{4}42​−41​=41​

STEP 5

The common denominator for 14\frac{1}{4}41​ and 18\frac{1}{8}81​ is 888. So we convert 14\frac{1}{4}41​ to 28\frac{2}{8}82​.14=28\frac{1}{4} = \frac{2}{8}41​=82​

STEP 6

Now add 18\frac{1}{8}81​ to 28\frac{2}{8}82​.28+18=38\frac{2}{8} + \frac{1}{8} = \frac{3}{8}82​+81​=83​

STEP 7

The common denominator for 3\frac{3}{}3​ and 116\frac{1}{16}161​ is 161616. So we convert 3\frac{3}{}3​ to 616\frac{6}{16}166​.3=616\frac{3}{} = \frac{6}{16}3​=166​

STEP 8

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QuestionCalcula la suma: $\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}=$

Assumptions1. We are working with real numbers. . The operations are basic arithmetic operations (addition and subtraction).
3. The fractions are all in the form of $\frac{1}{^n}$ where $n$ is a positive integer.

The common denominator for $\frac{1}{2}$ and $\frac{1}{}$ is $$. So we convert $\frac{1}{2}$ to $\frac{2}{}$.
$\frac{1}{2} = \frac{2}{}$

Now subtract $\frac{1}{4}$ from $\frac{2}{4}$ .
$\frac{2}{4} - \frac{1}{4} = \frac{1}{4}$

The common denominator for $\frac{1}{4}$ and $\frac{1}{8}$ is $8$ . So we convert $\frac{1}{4}$ to $\frac{2}{8}$ .
$\frac{1}{4} = \frac{2}{8}$

Now add $\frac{1}{8}$ to $\frac{2}{8}$ .
$\frac{2}{8} + \frac{1}{8} = \frac{3}{8}$

The common denominator for $\frac{3}{}$ and $\frac{1}{16}$ is $16$ . So we convert $\frac{3}{}$ to $\frac{6}{16}$ .
$\frac{3}{} = \frac{6}{16}$