The cubes of the natural numbers are grouped as `1^(3),(2^(3),3^(3)),(4^(3),5^(3),6^(3)), . . . . . . . .,` the the sum of the number in the `n^(th)` group, is

Options

A. `(1)/(8)n^(3)(n^(2)+1)(n^(2)+3)`
B. `(1)/(16)n^(3)(n^(2)+16)(n^(2)+12)`
C. `(n^(3))/(12)(n^(2)+2)(n^(2)+4)`
D. none of these

Answer

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

### Step 1: Identify the groups
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: \(7^3, 8^3, 9^3, 10^3\)
- And so on...

The number of terms in each group increases sequentially: the first group has 1 term, the second group has 2 terms, the third group has 3 terms, and so forth.

### Step 2: Determine the last term of the nth group
The last term of the nth group corresponds to the nth triangular number, which is given by the formula:
\[ T_n = \frac{n(n + 1)}{2} \]
This means the last term of the nth group is \(T_n^3\).

### Step 3: Determine the first term of the nth group
The first term of the nth group can be found by taking the last term of the previous group and adding 1:
\[ \text{First term of nth group} = T_{n-1} + 1 = \frac{(n-1)n}{2} + 1 \]

### Step 4: Count the number of terms in the nth group
The number of terms in the nth group is simply \(n\).

### Step 5: Write the terms in the nth group
The terms in the nth group can be expressed as:
\[ \left(\frac{(n-1)n}{2} + 1\right)^3, \left(\frac{(n-1)n}{2} + 2\right)^3, \ldots, \left(\frac{(n-1)n}{2} + n\right)^3 \]

### Step 6: Calculate the sum of the cubes in the nth group
The sum of the cubes in the nth group can be expressed as:
\[ \sum_{k=1}^{n} \left(\frac{(n-1)n}{2} + k\right)^3 \]

### Step 7: Use the formula for the sum of cubes
Using the formula for the sum of cubes, we know:
\[ \sum_{k=1}^{m} k^3 = \left(\frac{m(m + 1)}{2}\right)^2 \]
We can apply this to our expression to find the sum of the cubes in the nth group.

### Final Expression
The final expression for the sum of the cubes in the nth group is:
\[ \text{Sum} = \left(\frac{n(n + 1)}{2}\right)^2 - \left(\frac{(n-1)n}{2}\right)^2 \]
This simplifies to:
\[ \text{Sum} = \frac{1}{4} n^2 \left(n^2 + 2n + 1 - (n^2 - 2n + 1)\right) = \frac{1}{4} n^2 (4n) = n^2(n) \]

### Conclusion
Thus, the sum of the cubes of the natural numbers in the nth group is:
\[ \frac{n^2(n + 1)^2}{4} \]