The sum of the 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. 71
B. 96
C. 142
D. 192

Explanation

To find the sum of the first 9 terms of the series given by:

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

we first need to determine the general term \( T_n \) of the series.

### Step 1: Determine the nth term \( T_n \)

The nth term can be expressed as:

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

The numerator \( 1^3 + 2^3 + \ldots + n^3 \) is known to be:

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

The denominator \( 1 + 3 + 5 + \ldots + (2n - 1) \) is the sum of the first \( n \) odd numbers, which equals \( n^2 \).

Thus, we can write:

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

### Step 2: Simplify the nth term

Now, simplifying \( T_n \):

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

### Step 3: Calculate the sum of the first 9 terms

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

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

This can be factored out as:

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

### Step 4: Calculate \( \sum_{n=1}^{9} (n+1)^2 \)

We can rewrite \( (n+1)^2 \) as \( n^2 + 2n + 1 \):

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

Using the formulas for the sums:

- \( \sum_{n=1}^{k} n^2 = \frac{k(k+1)(2k+1)}{6} \)
- \( \sum_{n=1}^{k} n = \frac{k(k+1)}{2} \)

For \( k = 9 \):

\[ \sum_{n=1}^{9} n^2 = \frac{9 \cdot 10 \cdot 19}{6} = 285 \]

\[ \sum_{n=1}^{9} n = \frac{9 \cdot 10}{2} = 45 \]

\[ \sum_{n=1}^{9} 1 = 9 \]

Now substituting back:

\[ \sum_{n=1}^{9} (n+1)^2 = 285 + 2 \cdot 45 + 9 = 285 + 90 + 9 = 384 \]

### Step 5: Final Calculation of \( S_9 \)

Now, substituting back into \( S_9 \):

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

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

\[ \boxed{96} \]