`(3)/(4)-(5)/(4^(2))+(3)/(4^(3))-(5)/(4^(4))+(3)/(4^(5))-(5)/(4^(6))+.....oo=`?

Options

A. `(8)/(15)`
B. `(7)/(17)`
C. `(7)/(15)`
D. `(8)/(17)`

Answer

To solve the series

\[ \frac{3}{4} - \frac{5}{4^2} + \frac{3}{4^3} - \frac{5}{4^4} + \frac{3}{4^5} - \frac{5}{4^6} + \ldots \]

we can rearrange the terms to group the positive and negative terms separately.

### Step 1: Rearrange the Series

We can separate the series into two parts:

1. Positive terms:
\[ \frac{3}{4} + \frac{3}{4^3} + \frac{3}{4^5} + \ldots \]

2. Negative terms:
\[ -\left(\frac{5}{4^2} + \frac{5}{4^4} + \frac{5}{4^6} + \ldots\right) \]

### Step 2: Identify the First Series

The first series (positive terms) can be expressed as:
\[ \frac{3}{4} \left(1 + \frac{1}{4^2} + \frac{1}{4^4} + \ldots\right) \]

This is a geometric series where \(a = 1\) and \(r = \frac{1}{16}\).

### Step 3: Calculate the Sum of the First Series

The sum \(S_1\) of an infinite geometric series is given by:
\[ S = \frac{a}{1 - r} \]
Thus, for our series:
\[ S_1 = \frac{1}{1 - \frac{1}{16}} = \frac{1}{\frac{15}{16}} = \frac{16}{15} \]

Now, substituting back:
\[ S_1 = \frac{3}{4} \cdot \frac{16}{15} = \frac{48}{60} = \frac{4}{5} \]

### Step 4: Identify the Second Series

The second series (negative terms) can be expressed as:
\[ -\frac{5}{4^2} \left(1 + \frac{1}{4^2} + \frac{1}{4^4} + \ldots\right) \]

This is also a geometric series where \(a = 1\) and \(r = \frac{1}{16}\).

### Step 5: Calculate the Sum of the Second Series

Using the same formula for the sum:
\[ S_2 = \frac{1}{1 - \frac{1}{16}} = \frac{16}{15} \]

Now substituting back:
\[ S_2 = -\frac{5}{16} \cdot \frac{16}{15} = -\frac{5}{15} = -\frac{1}{3} \]

### Step 6: Combine the Two Sums

Now we combine both sums:
\[ S = S_1 + S_2 = \frac{4}{5} - \frac{1}{3} \]

### Step 7: Find a Common Denominator

To subtract these fractions, we need a common denominator. The LCM of 5 and 3 is 15.

Rewriting the fractions:
\[ \frac{4}{5} = \frac{12}{15}, \quad \text{and} \quad \frac{1}{3} = \frac{5}{15} \]

### Step 8: Perform the Subtraction

Now we can perform the subtraction:
\[ S = \frac{12}{15} - \frac{5}{15} = \frac{7}{15} \]

### Final Answer

Thus, the value of the series is:
\[ \boxed{\frac{7}{15}} \]