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solve everything and start from 25 b, detaily!

Let's start with **Problem 25(b)**, which asks us to prove the following identity by induction:

\[
\sum_{k=0}^n k^3 = \left(\frac{n(n+1)}{2}\right)^2.
\]

### Step 1: Base Case ( \( n = 0 \) )

For \( n = 0 \), the sum on the left-hand side is:
\[
\sum_{k=0}^0 k^3 = 0^3 = 0.
\]

The right-hand side for \( n = 0 \) is:
\[
\left(\frac{0 \cdot (0+1)}{2}\right)^2 = \left(\frac{0}{2}\right)^2 = 0.
\]

Thus, both sides are equal for \( n = 0 \). The base case holds.

### Step 2: Inductive Hypothesis

Assume that the formula holds for some \( n = m \); that is, assume:
\[
\sum_{k=0}^m k^3 = \left(\frac{m(m+1)}{2}\right)^2.
\]

We need to show that this implies the formula also holds for \( n = m + 1 \).

### Step 3: Inductive Step

For \( n = m+1 \), the sum can be expanded as:
\[
\sum_{k=0}^{m+1} k^3 = \sum_{k=0}^m k^3 + (m+1)^3.
\]

By the inductive hypothesis, we can substitute \( \sum_{k=0}^m k^3 \) with \( \left(\frac{m(m+1)}{2}\right)^2 \). Therefore:
\[
\sum_{k=0}^{m+1} k^3 = \left(\frac{m(m+1)}{2}\right)^2 + (m+1)^3.
\]

We need to show that this expression is equal to \( \left(\frac{(m+1)(m+2)}{2}\right)^2 \).

#### Expanding the Right-Hand Side for \( n = m+1 \)

The right-hand side for \( n = m+1 \) is:
\[
\left(\frac{(m+1)(m+2)}{2}\right)^2 = \frac{(m+1)^2 (m+2)^2}{4}.
\]

#### Simplifying the Left-Hand Side

Let's expand and simplify the left-hand side expression:
\[
\left(\frac{m(m+1)}{2}\right)^2 + (m+1)^3 = \frac{m^2(m+1)^2}{4} + (m+1)^3.
\]

Factor \( (m+1)^2 \) from both terms:
\[
= (m+1)^2 \left(\frac{m^2}{4} + (m+1)\right).
\]

Now, combine the terms inside the parentheses:
\[
= (m+1)^2 \left(\frac{m^2 + 4m + 4}{4}\right).
\]

This simplifies to:
\[
= (m+1)^2 \cdot \frac{(m+2)^2}{4} = \frac{(m+1)^2 (m+2)^2}{4}.
\]

#### Conclusion of Inductive Step

We have shown that:
\[
\sum_{k=0}^{m+1} k^3 = \frac{(m+1)^2 (m+2)^2}{4} = \left(\frac{(m+1)(m+2)}{2}\right)^2.
\]

Thus, by the principle of mathematical induction, the formula holds for all \( n \in \mathbb{N} \).

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Would you like to proceed with the next problem in the set, or do you have any questions on this solution?

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Here are five questions to deepen your understanding:

1. How does mathematical induction work, and why is it used here?
2. Can this method be used to prove identities involving higher powers, like \( k^4 \)?
3. Why is factoring out \( (m+1)^2 \) helpful in the inductive step?
4. What are some practical applications of summing cubes in mathematics?
5. How does this sum relate to other formulas in number theory?

**Tip:** For complex identities, verifying base cases and step-by-step simplification in the induction step can help ensure accuracy.

Solve everything and start from 25(b), detaily!

Learn how to prove the formula \( \sum_{k=0}^n k^3 = \left(\frac{n(n+1)}{2}\right)^2 \) using mathematical induction. Detailed steps are provided for the base case, inductive hypothesis, and inductive step to guide you through the solution.

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