The cubes of the natural numbers are grouped as `1^3 , (2^3, 3^3) , (4^3 , 5^3 , 6^3) `, … then the sum of the numbers in the nth group is :

Options

A. a. `(n^3)/(12) (n^2 + 2) (n^2 + 4)`
B. b. `1/8 n^3 (n^2 +1) (n^2 + 3)`
C. c. `(n^3)/(6) (n^2 + 1) (n^2 + 2) `
D. d. none of these

Answer

To solve the problem of finding the sum of the cubes of natural numbers grouped as described, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Pattern of Grouping:**
The natural numbers are grouped as follows:
- Group 1: \(1^3\)
- Group 2: \(2^3, 3^3\)
- Group 3: \(4^3, 5^3, 6^3\)
- Group 4: \(7^3, 8^3, 9^3, 10^3\)
- ...

We can see that the number of terms in each group increases by 1 for each subsequent group.

2. **Determine the Last Term in the nth Group:**
The last term in the nth group can be represented as:
\[ T_n = \frac{n(n + 1)}{2} \]
This formula gives the total number of terms up to the nth group.

3. **Calculate the First Term of the nth Group:**
The first term of the nth group can be calculated as:
\[ \text{First term} = T_{n-1} + 1 = \frac{(n-1)n}{2} + 1 \]

4. **List the Terms in the nth Group:**
The terms in the nth group will be:
\[ \left(\frac{(n-1)n}{2} + 1\right)^3, \left(\frac{(n-1)n}{2} + 2\right)^3, \ldots, \left(\frac{(n-1)n}{2} + n\right)^3 \]

5. **Sum of the Cubes in the nth Group:**
The sum of the cubes of the first n natural numbers is given by the formula:
\[ \left(\frac{n(n + 1)}{2}\right)^2 \]
Therefore, the sum of the cubes in the nth group can be expressed as:
\[ S_n = \sum_{k=1}^{n} \left(\frac{(n-1)n}{2} + k\right)^3 \]

6. **Using the Formula for the Sum of Cubes:**
We can simplify the sum using the formula for the sum of cubes:
\[ S_n = \left(\frac{n(n + 1)}{2}\right)^2 - \left(\frac{(n-1)n}{2}\right)^2 \]

7. **Final Expression:**
After simplification, we find that the sum of the cubes in the nth group is:
\[ S_n = \frac{n^2(n + 1)^2}{4} - \frac{(n-1)^2n^2}{4} \]
This simplifies to:
\[ S_n = \frac{n^2(n + 1)(n + 1 - n + 1)}{4} = \frac{n^2(n + 1)}{4} \]

### Conclusion:
The sum of the numbers in the nth group is:
\[ S_n = \frac{n^2(n + 1)}{4} \]