Sum up to 16 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. 450
B. 456
C. 446
D. None of these

Explanation

To find the sum of the first 16 terms of the series

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

we will denote the \( n \)-th term of the series as \( T_n \).

### Step 1: Identify the \( n \)-th term \( T_n \)

The numerator of \( T_n \) is the sum of cubes from 1 to \( n \):
\[ T_n = \frac{1^3 + 2^3 + 3^3 + \ldots + n^3}{1 + 3 + 5 + \ldots + (2n - 1)} \]

The denominator is the sum of the first \( n \) odd numbers, which can be expressed as:
\[ 1 + 3 + 5 + \ldots + (2n - 1) = n^2 \]

Thus, we can write:
\[ T_n = \frac{\sum_{k=1}^{n} k^3}{n^2} \]

### Step 2: Use the formula for the sum of cubes

The formula for the sum of the first \( n \) cubes is:
\[ \sum_{k=1}^{n} k^3 = \left(\frac{n(n + 1)}{2}\right)^2 \]

So we can substitute this into our expression for \( T_n \):
\[ T_n = \frac{\left(\frac{n(n + 1)}{2}\right)^2}{n^2} \]

### Step 3: Simplify \( T_n \)

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

### Step 4: Find the sum of the first 16 terms

Now we need to find the sum of the first 16 terms, denoted as \( S_{16} \):
\[ S_{16} = T_1 + T_2 + T_3 + \ldots + T_{16} \]

Substituting our expression for \( T_n \):
\[ S_{16} = \sum_{n=1}^{16} T_n = \sum_{n=1}^{16} \frac{(n + 1)^2}{4} \]

This can be factored out:
\[ S_{16} = \frac{1}{4} \sum_{n=1}^{16} (n + 1)^2 \]

### Step 5: Calculate \( \sum_{n=1}^{16} (n + 1)^2 \)

We can rewrite \( (n + 1)^2 \) as \( n^2 + 2n + 1 \):
\[ \sum_{n=1}^{16} (n + 1)^2 = \sum_{n=1}^{16} (n^2 + 2n + 1) = \sum_{n=1}^{16} n^2 + 2\sum_{n=1}^{16} n + \sum_{n=1}^{16} 1 \]

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

For \( k = 16 \):
\[ \sum_{n=1}^{16} n^2 = \frac{16 \cdot 17 \cdot 33}{6} = 1496 \]
\[ \sum_{n=1}^{16} n = \frac{16 \cdot 17}{2} = 136 \]
\[ \sum_{n=1}^{16} 1 = 16 \]

Putting it all together:
\[ \sum_{n=1}^{16} (n + 1)^2 = 1496 + 2 \cdot 136 + 16 = 1496 + 272 + 16 = 1784 \]

### Step 6: Calculate \( S_{16} \)

Now substituting back:
\[ S_{16} = \frac{1}{4} \cdot 1784 = 446 \]

### Final Answer

Thus, the sum of the first 16 terms of the series is:
\[ \boxed{446} \]