Certainly! Let’s go through Method 2 step by step to show that $\sum_{i=1}^{n} i^3 = \left(\frac{n(n+1)}{2}\right)^2$.

Step 1: Consider the Area of the Square

Consider a square where each side is divided into segments of lengths $1, 2, 3, \ldots, n$. The total length of each side is: $\text{Side length} = 1 + 2 + 3 + \cdots + n = \frac{n(n+1)}{2}$ So, the area $A$ of this square is: $A = \left(\frac{n(n+1)}{2}\right)^2$

Step 2: Decomposing the Square into Smaller Areas

Now, let's decompose this square into smaller areas $a_1, a_2, \ldots, a_n$:

• $a_1$ is a unit square with area $1^2 = 1$.
• $a_2, a_3, \ldots, a_n$ are areas formed by squares with side lengths equal to $i = 2, 3, \ldots, n$.

The area $a_i$ of the $i$th square is given by the volume formed by summing up squares along the diagonal and the addition of the segments that connect these squares in the grid. For each $i$, the area $a_i$ will be: $a_i = i^3$

This is because each segment forms a cube, and the volume (or in this case, area of the shape in the plane of the square) generated by the length $i$ contributes $i^3$ to the total area.

Step 3: Summing the Areas

The total area of the square is also the sum of these smaller areas: $A = a_1 + a_2 + \cdots + a_n = \sum_{i=1}^{n} i^3$

Step 4: Equating the Two Expressions for the Area

We already calculated that the area of the square is $A = \left(\frac{n(n+1)}{2}\right)^2$, and we also found that the same area is $A = \sum_{i=1}^{n} i^3$.

Thus, equating these two expressions, we get: $\sum_{i=1}^{n} i^3 = \left(\frac{n(n+1)}{2}\right)^2$

This completes the proof using Method 2.

Do you want details or have any questions?

Here are some related questions:

1. How does the concept of a telescoping sum work in Method 1?
2. Can you provide an example with a small $n$ (like $n=3$) to see the pattern of the sums?
3. How can we visualize this geometric method for larger $n$?
4. Why is the side length of the square equal to the sum of the first $n$ integers?
5. How can this method be generalized to other power sums, like $\sum_{i=1}^{n} i^2$?

Tip: Visualizing mathematical concepts, especially with geometric interpretations like this one, can often make abstract ideas more tangible and easier to understand.