If the sum of the series `1+(3)/(2)+(5)/(4)+(7)/(8)+……+((2n-1))/((2)^(n-1))` is `f(n)`, then the value of `f(8)` is

Options

A. `4+(12)/(2^(5))`
B. `5+(13)/(2^(7))`
C. `6-(19)/(2^(7))`
D. `5-(13)/(2^(7))`

Answer

To solve the problem, we need to find the sum of the series given by:

\[ f(n) = 1 + \frac{3}{2} + \frac{5}{4} + \frac{7}{8} + \ldots + \frac{2n-1}{2^{n-1}} \]

We will denote this sum as \( S_n \).

### Step 1: Write the series
The series can be expressed as:

\[ S_n = \sum_{k=1}^{n} \frac{2k-1}{2^{k-1}} \]

### Step 2: Multiply the series by \( \frac{1}{2} \)
Now, we multiply the entire series \( S_n \) by \( \frac{1}{2} \):

\[ \frac{1}{2} S_n = \frac{1}{2} + \frac{3}{4} + \frac{5}{8} + \ldots + \frac{2n-1}{2^n} \]

### Step 3: Shift the terms
Notice that if we shift the terms of \( \frac{1}{2} S_n \) one place to the right, we can write:

\[ \frac{1}{2} S_n = \frac{1}{2} + \frac{3}{4} + \frac{5}{8} + \ldots + \frac{2(n-1)-1}{2^{n-1}} + \frac{2n-1}{2^n} \]

### Step 4: Subtract the two series
Now we subtract \( \frac{1}{2} S_n \) from \( S_n \):

\[ S_n - \frac{1}{2} S_n = 1 + \left( \frac{3}{2} - \frac{1}{2} \right) + \left( \frac{5}{4} - \frac{3}{4} \right) + \ldots + \left( \frac{2n-1}{2^{n-1}} - \frac{2(n-1)-1}{2^n} \right) \]

This simplifies to:

\[ \frac{1}{2} S_n = 1 + \frac{2}{2} + \frac{2}{4} + \frac{2}{8} + \ldots + \frac{2}{2^{n-1}} - \frac{2n-1}{2^n} \]

### Step 5: Recognize the geometric series
The series \( \frac{2}{2} + \frac{2}{4} + \frac{2}{8} + \ldots + \frac{2}{2^{n-1}} \) is a geometric series with first term \( \frac{2}{2} = 1 \) and common ratio \( \frac{1}{2} \):

\[ \text{Sum} = 1 \cdot \frac{1 - \left( \frac{1}{2} \right)^{n-1}}{1 - \frac{1}{2}} = 2 \left( 1 - \frac{1}{2^{n-1}} \right) \]

### Step 6: Substitute back into the equation
Now substituting back, we have:

\[ \frac{1}{2} S_n = 1 + 2 \left( 1 - \frac{1}{2^{n-1}} \right) - \frac{2n-1}{2^n} \]

### Step 7: Solve for \( S_n \)
Multiplying through by 2 gives:

\[ S_n = 2 + 4 \left( 1 - \frac{1}{2^{n-1}} \right) - \frac{2(2n-1)}{2^n} \]

Simplifying further:

\[ S_n = 6 - \frac{4}{2^{n-1}} - \frac{2n-1}{2^{n-1}} = 6 - \frac{2n + 3}{2^{n-1}} \]

### Step 8: Find \( f(8) \)
Now we substitute \( n = 8 \):

\[ f(8) = 6 - \frac{2(8) + 3}{2^{8-1}} = 6 - \frac{19}{128} \]

### Step 9: Combine terms
To combine:

\[ f(8) = 6 - \frac{19}{128} = \frac{768 - 19}{128} = \frac{749}{128} \]

### Final Answer
Thus, the value of \( f(8) \) is:

\[ f(8) = 6 - \frac{19}{128} \]