`sum_(r=1)^(10)(r+{(r)/(7)}+{(2r)/(7)})=` where `{x} rarr` fractional part function

Options

A. `(445)/(7)`
B. `(454)/(7)`
C. `(447)/(17)`
D. `(447)/(7)`

Explanation

To solve the given summation \( \sum_{r=1}^{10} \left( r + \{ \frac{r}{7} \} + \{ \frac{2r}{7} \} \right) \), we can break it down into three separate summations:

1. \( \sum_{r=1}^{10} r \)
2. \( \sum_{r=1}^{10} \{ \frac{r}{7} \} \)
3. \( \sum_{r=1}^{10} \{ \frac{2r}{7} \} \)

### Step 1: Calculate \( \sum_{r=1}^{10} r \)

The sum of the first \( n \) natural numbers is given by the formula:

\[ \sum_{r=1}^{n} r = \frac{n(n+1)}{2} \]

For \( n = 10 \):

\[ \sum_{r=1}^{10} r = \frac{10(10+1)}{2} = \frac{10 \times 11}{2} = 55 \]

### Step 2: Calculate \( \sum_{r=1}^{10} \{ \frac{r}{7} \} \)

The fractional part function \( \{ x \} = x - \lfloor x \rfloor \). We need to find \( \{ \frac{r}{7} \} \) for \( r = 1, 2, \ldots, 10 \):

- For \( r = 1 \): \( \{ \frac{1}{7} \} = \frac{1}{7} \)
- For \( r = 2 \): \( \{ \frac{2}{7} \} = \frac{2}{7} \)
- For \( r = 3 \): \( \{ \frac{3}{7} \} = \frac{3}{7} \)
- For \( r = 4 \): \( \{ \frac{4}{7} \} = \frac{4}{7} \)
- For \( r = 5 \): \( \{ \frac{5}{7} \} = \frac{5}{7} \)
- For \( r = 6 \): \( \{ \frac{6}{7} \} = \frac{6}{7} \)
- For \( r = 7 \): \( \{ \frac{7}{7} \} = 0 \)
- For \( r = 8 \): \( \{ \frac{8}{7} \} = \frac{1}{7} \)
- For \( r = 9 \): \( \{ \frac{9}{7} \} = \frac{2}{7} \)
- For \( r = 10 \): \( \{ \frac{10}{7} \} = \frac{3}{7} \)

Now, summing these values:

\[ \sum_{r=1}^{10} \{ \frac{r}{7} \} = \frac{1}{7} + \frac{2}{7} + \frac{3}{7} + \frac{4}{7} + \frac{5}{7} + \frac{6}{7} + 0 + \frac{1}{7} + \frac{2}{7} + \frac{3}{7} \]

This simplifies to:

\[ = \frac{1 + 2 + 3 + 4 + 5 + 6 + 0 + 1 + 2 + 3}{7} = \frac{27}{7} \]

### Step 3: Calculate \( \sum_{r=1}^{10} \{ \frac{2r}{7} \} \)

Now we find \( \{ \frac{2r}{7} \} \):

- For \( r = 1 \): \( \{ \frac{2}{7} \} = \frac{2}{7} \)
- For \( r = 2 \): \( \{ \frac{4}{7} \} = \frac{4}{7} \)
- For \( r = 3 \): \( \{ \frac{6}{7} \} = \frac{6}{7} \)
- For \( r = 4 \): \( \{ \frac{8}{7} \} = \frac{1}{7} \)
- For \( r = 5 \): \( \{ \frac{10}{7} \} = \frac{3}{7} \)
- For \( r = 6 \): \( \{ \frac{12}{7} \} = \frac{5}{7} \)
- For \( r = 7 \): \( \{ \frac{14}{7} \} = 0 \)
- For \( r = 8 \): \( \{ \frac{16}{7} \} = \frac{2}{7} \)
- For \( r = 9 \): \( \{ \frac{18}{7} \} = \frac{4}{7} \)
- For \( r = 10 \): \( \{ \frac{20}{7} \} = \frac{6}{7} \)

Summing these values:

\[ \sum_{r=1}^{10} \{ \frac{2r}{7} \} = \frac{2}{7} + \frac{4}{7} + \frac{6}{7} + \frac{1}{7} + \frac{3}{7} + \frac{5}{7} + 0 + \frac{2}{7} + \frac{4}{7} + \frac{6}{7} \]

This simplifies to:

\[ = \frac{2 + 4 + 6 + 1 + 3 + 5 + 0 + 2 + 4 + 6}{7} = \frac{33}{7} \]

### Final Calculation

Now we combine all three parts:

\[ \sum_{r=1}^{10} \left( r + \{ \frac{r}{7} \} + \{ \frac{2r}{7} \} \right) = 55 + \frac{27}{7} + \frac{33}{7} \]

Combine the fractional parts:

\[ = 55 + \frac{27 + 33}{7} = 55 + \frac{60}{7} = 55 + 8.57 \approx 63.57 \]

To express this as a single fraction:

\[ = \frac{55 \times 7 + 60}{7} = \frac{385 + 60}{7} = \frac{445}{7} \]

Thus, the final answer is:

\[ \sum_{r=1}^{10} \left( r + \{ \frac{r}{7} \} + \{ \frac{2r}{7} \} \right) = \frac{445}{7} \]