Value of :`1^3+2^3+3^3++n^3=`

Options

A. `{(n(n+1))/2}^2dot`
B. `{(n(n+1))/2}^3dot`
C. `{(n(n+1))/2}dot`
D. None of these

Explanation

To find the value of the sum \( S = 1^3 + 2^3 + 3^3 + \ldots + n^3 \), we can use the formula for the sum of cubes of the first \( n \) natural numbers. The formula is:

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

### Step-by-Step Solution:

1. **Understanding the Problem:**
We need to find the sum of cubes from \( 1^3 \) to \( n^3 \).

2. **Using the Formula:**
The formula for the sum of the first \( n \) cubes is given by:
\[ S = \left( \frac{n(n+1)}{2} \right)^2 \]

3. **Substituting the Value of \( n \):**
For any specific value of \( n \), you can substitute it into the formula. For example, if \( n = 3 \):
\[ S = \left( \frac{3(3+1)}{2} \right)^2 = \left( \frac{3 \times 4}{2} \right)^2 = \left( \frac{12}{2} \right)^2 = 6^2 = 36 \]

4. **General Case:**
Thus, for any \( n \), the sum of cubes can be calculated as:
\[ S = \left( \frac{n(n+1)}{2} \right)^2 \]

5. **Conclusion:**
Therefore, the value of \( 1^3 + 2^3 + 3^3 + \ldots + n^3 \) is:
\[ \left( \frac{n(n+1)}{2} \right)^2 \]