The sum of first 9 terms of the series `(1^(3))/(1)+(1^(3)+2^(3))/(1+3)+(1^(3)+2^(3)+3^(3))/(1+3+5)+"........"` is

Options

A. 192
B. 71
C. 96
D. 142

Explanation

To find the sum of the first 9 terms of the given series, we will break down the problem step by step.

### Step 1: Understanding the series
The series is given as:
\[ \frac{1^3}{1} + \frac{1^3 + 2^3}{1 + 3} + \frac{1^3 + 2^3 + 3^3}{1 + 3 + 5} + \ldots \]
The n-th term of the series can be expressed as:
\[ T_n = \frac{1^3 + 2^3 + \ldots + n^3}{1 + 3 + 5 + \ldots + (2n - 1)} \]

### Step 2: Finding the numerator
The sum of the cubes of the first n natural numbers is given by the formula:
\[ 1^3 + 2^3 + \ldots + n^3 = \left(\frac{n(n+1)}{2}\right)^2 \]

### Step 3: Finding the denominator
The sum of the first n odd numbers is:
\[ 1 + 3 + 5 + \ldots + (2n - 1) = n^2 \]

### Step 4: Expressing T_n
Now we can express \( T_n \):
\[ T_n = \frac{\left(\frac{n(n+1)}{2}\right)^2}{n^2} = \frac{n^2(n+1)^2}{4n^2} = \frac{(n+1)^2}{4} \]

### Step 5: Finding the first 9 terms
Now we need to find the sum of the first 9 terms:
\[ S_9 = T_1 + T_2 + T_3 + \ldots + T_9 \]
Substituting \( n = 1, 2, \ldots, 9 \):
\[ S_9 = \sum_{n=1}^{9} T_n = \sum_{n=1}^{9} \frac{(n+1)^2}{4} \]
This can be simplified as:
\[ S_9 = \frac{1}{4} \sum_{n=1}^{9} (n+1)^2 = \frac{1}{4} \sum_{n=2}^{10} n^2 \]

### Step 6: Calculating the sum of squares
The sum of squares of the first m natural numbers is given by:
\[ \sum_{k=1}^{m} k^2 = \frac{m(m+1)(2m+1)}{6} \]
For \( m = 10 \):
\[ \sum_{k=1}^{10} k^2 = \frac{10 \cdot 11 \cdot 21}{6} = 385 \]
Thus,
\[ \sum_{n=2}^{10} n^2 = 385 - 1^2 = 385 - 1 = 384 \]

### Step 7: Final calculation
Now substituting back into \( S_9 \):
\[ S_9 = \frac{1}{4} \cdot 384 = 96 \]

### Conclusion
The sum of the first 9 terms of the series is:
\[ \boxed{96} \]