To prove the formula:

$\sum_{k=1}^n k^3 = \frac{n^2 (n+1)^2}{4}$

we will use a well-known formula for the sum of the cubes of the first $n$ natural numbers.

Proof Outline

1. Recall the formula: The formula for the sum of cubes of the first $n$ natural numbers is: $\sum_{k=1}^n k^3 = \left( \sum_{k=1}^n k \right)^2$ where $\sum_{k=1}^n k = \frac{n(n+1)}{2}$.

2. Square the sum of integers: Using $\sum_{k=1}^n k = \frac{n(n+1)}{2}$, we can substitute this into the square: $\sum_{k=1}^n k^3 = \left( \frac{n(n+1)}{2} \right)^2$

3. Simplify the expression: Expanding the square gives: $\sum_{k=1}^n k^3 = \frac{n^2 (n+1)^2}{4}$

Thus, we have shown that:

$\sum_{k=1}^n k^3 = \frac{n^2 (n+1)^2}{4}$

This completes the proof.

Would you like a more detailed explanation of any steps?

Related Questions

1. How can we derive the formula for the sum of the first $n$ natural numbers?
2. What is the formula for the sum of squares and how is it derived?
3. Can we generalize this approach to find sums of higher powers, like $k^4$?
4. How is this formula useful in calculus or other fields of mathematics?
5. What other techniques can we use to prove series identities?

Tip:

Remember that knowing the basic formulas for sums of powers, such as squares and cubes, can be very helpful in solving various types of series problems efficiently.