"Show that nP
i=1
i3 =  n(n+1)
2
2
using two methods.
method 1 Use a telescoping sum as in class.
method 2 Consider a square where each side has
segments of length 1, 2, . . . , n; then the side length
is 1 + 2 + · · · + n = n(n+1)
2 and the area of the square
is A =  n(n+1)
2
2
. Now consider areas a1, a2, . . . , an,
where a1 is a unit square and a2, a3, . . . , an are as
shown in the figure. Show that ai = i3 for i =
1, . . . , n. Hence the area of the square is also equal
to A = a1 + a2 + · · · + an = nP
i=1
ai = nP
i=1
i3"
can you do method 2

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.

Explore two methods to prove that the sum of cubes of the first n natural numbers equals the square of the sum using geometric and algebraic approaches. Understand the relationship between the areas of a square and the volumes of cubes formed by segments along its sides. Suitable for advanced mathematics students.

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