If the sum of the first 15 terms of the series `((3)/(4))^(3)+(1(1)/(2))^(3)+(2(1)/(4))^(3)+3^(3)+(3(3)/(4))^(3)+...` is equal to 225k, then k is equal to

Options

A. 9
B. 27
C. 108
D. 54

Answer

To find the value of \( k \) in the equation where the sum of the first 15 terms of the series is equal to \( 225k \), we need to analyze the series given.

The series is:
\[ \left(\frac{3}{4}\right)^3 + \left(\frac{1}{2}\right)^3 + \left(\frac{2}{4}\right)^3 + 3^3 + \left(\frac{3}{4}\right)^3 + \ldots \]

### Step 1: Identify the pattern in the series

From the terms provided, we can see that the series consists of cubes of fractions and integers. The first few terms are:
1. \( \left(\frac{3}{4}\right)^3 \)
2. \( \left(\frac{1}{2}\right)^3 \)
3. \( \left(\frac{2}{4}\right)^3 = \left(\frac{1}{2}\right)^3 \)
4. \( 3^3 \)
5. \( \left(\frac{3}{4}\right)^3 \)
6. \( \ldots \)

It appears that the terms are repeating, and we need to determine how many unique terms there are and their contributions to the sum.

### Step 2: Calculate the first few unique terms

Let's calculate the cubes of the unique terms:
1. \( \left(\frac{3}{4}\right)^3 = \frac{27}{64} \)
2. \( \left(\frac{1}{2}\right)^3 = \frac{1}{8} = \frac{8}{64} \)
3. \( 3^3 = 27 \)

### Step 3: Sum the unique terms

Now we can sum these unique terms:
\[ \text{Sum of unique terms} = \left(\frac{27}{64}\right) + \left(\frac{8}{64}\right) + 27 \]
Converting \( 27 \) to a fraction with a denominator of \( 64 \):
\[ 27 = \frac{27 \times 64}{64} = \frac{1728}{64} \]
Now, summing these:
\[ \text{Sum} = \frac{27 + 8 + 1728}{64} = \frac{1763}{64} \]

### Step 4: Determine the number of terms

Since the series is repeating, we need to find out how many times each term appears in the first 15 terms.

Assuming the pattern repeats every 3 terms:
- The first term \( \left(\frac{3}{4}\right)^3 \) appears 5 times,
- The second term \( \left(\frac{1}{2}\right)^3 \) appears 5 times,
- The third term \( 3^3 \) appears 5 times.

### Step 5: Calculate the total sum of the first 15 terms

Now we calculate the total sum of the first 15 terms:
\[ \text{Total Sum} = 5 \times \left(\frac{27}{64}\right) + 5 \times \left(\frac{8}{64}\right) + 5 \times 27 \]
Calculating each part:
\[ = \frac{135}{64} + \frac{40}{64} + 135 \]
Converting \( 135 \) to a fraction:
\[ 135 = \frac{135 \times 64}{64} = \frac{8640}{64} \]
Now summing:
\[ \text{Total Sum} = \frac{135 + 40 + 8640}{64} = \frac{8815}{64} \]

### Step 6: Set the total sum equal to \( 225k \)

We know from the problem statement that this sum equals \( 225k \):
\[ \frac{8815}{64} = 225k \]

### Step 7: Solve for \( k \)

To find \( k \):
\[ k = \frac{8815}{64 \times 225} \]
Calculating \( 64 \times 225 = 14400 \):
\[ k = \frac{8815}{14400} \]

### Step 8: Simplify \( k \)

To simplify \( \frac{8815}{14400} \), we can check for common factors. The GCD of 8815 and 14400 is 5:
\[ k = \frac{8815 \div 5}{14400 \div 5} = \frac{1763}{2880} \]

Thus, the final answer is:
\[ \boxed{\frac{1763}{2880}} \]