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 series given by

\[ \frac{1^3}{1} + \frac{1^3 + 2^3}{1 + 3} + \frac{1^3 + 2^3 + 3^3}{1 + 3 + 5} + \ldots \]

we need to determine the general term of the series.

### Step 1: Identify the nth term of the series
The nth term \( T_n \) can be expressed as:

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

### Step 2: Simplify the numerator
The sum of the cubes of the first n natural numbers is given by the formula:

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

### Step 3: Simplify the denominator
The sum of the first n odd numbers is:

\[ 1 + 3 + 5 + \ldots + (2n - 1) = n^2 \]

### Step 4: Write the nth term
Substituting the simplified forms into the nth term, we have:

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

### Step 5: Find the sum of the first 9 terms
Now we need to find \( S_9 \), the sum of the first 9 terms:

\[ S_9 = \sum_{n=1}^{9} T_n = \sum_{n=1}^{9} \frac{(n + 1)^2}{4} \]

### Step 6: Factor out the constant
Factoring out \( \frac{1}{4} \):

\[ S_9 = \frac{1}{4} \sum_{n=1}^{9} (n + 1)^2 \]

### Step 7: Calculate the sum \( \sum_{n=1}^{9} (n + 1)^2 \)
This can be rewritten as:

\[ \sum_{n=1}^{9} (n + 1)^2 = \sum_{n=2}^{10} n^2 \]

Using the formula for the sum of squares:

\[ \sum_{k=1}^{m} k^2 = \frac{m(m + 1)(2m + 1)}{6} \]

For \( m = 10 \):

\[ \sum_{k=1}^{10} k^2 = \frac{10 \cdot 11 \cdot 21}{6} = 385 \]

For \( m = 1 \):

\[ \sum_{k=1}^{1} k^2 = 1 \]

Thus,

\[ \sum_{n=2}^{10} n^2 = 385 - 1 = 384 \]

### Step 8: Calculate \( S_9 \)
Now substituting back:

\[ S_9 = \frac{1}{4} \cdot 384 = 96 \]

Thus, the sum of the first 9 terms of the series is:

\[ \boxed{96} \]