Certainly! Let's break down the proof for Problem 1a in more detail, step by step.

Problem 1a:

Statement:
Prove by mathematical induction that: $1^2 + 2^2 + 3^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6}$

Step 1: Base Case

We start by verifying the formula for the smallest value of $n$, which is $n = 1$.

For $n = 1$: $1^2 = 1$

According to the formula: $\frac{1(1+1)(2(1)+1)}{6} = \frac{1 \times 2 \times 3}{6} = \frac{6}{6} = 1$

Since both sides of the equation are equal, the base case holds true.

Step 2: Inductive Hypothesis

Next, we assume that the formula is true for some arbitrary positive integer $k$. This assumption is known as the inductive hypothesis.

Assume: $1^2 + 2^2 + 3^2 + \cdots + k^2 = \frac{k(k+1)(2k+1)}{6}$

We assume this is true and use it to prove that the formula also holds for $n = k+1$.

Step 3: Inductive Step

We need to prove that if the formula is true for $n = k$, then it must also be true for $n = k+1$.

We need to show: $1^2 + 2^2 + 3^2 + \cdots + k^2 + (k+1)^2 = \frac{(k+1)((k+1)+1)(2(k+1)+1)}{6}$

Using the inductive hypothesis, we substitute the sum up to $k^2$: $\frac{k(k+1)(2k+1)}{6} + (k+1)^2$

Now, let's work on simplifying the right-hand side:

1. Factor out $(k+1)$ from both terms: $= \frac{k(k+1)(2k+1)}{6} + \frac{6(k+1)^2}{6}$ $= \frac{(k+1)}{6} \left[ k(2k+1) + 6(k+1) \right]$

2. Expand and combine the terms inside the bracket: $= \frac{(k+1)}{6} \left[ 2k^2 + k + 6k + 6 \right]$ $= \frac{(k+1)}{6} \left[ 2k^2 + 7k + 6 \right]$

3. Factorize the quadratic expression $2k^2 + 7k + 6$: $= \frac{(k+1)}{6} \times (2k + 3)(k + 2)$

4. Rewrite the expression: $= \frac{(k+1)(k+2)(2k+3)}{6}$

This expression matches the form we need to prove: $\frac{(k+1)((k+1)+1)(2(k+1)+1)}{6}$

Step 4: Conclusion

Since we've shown that if the formula is true for $n = k$, then it must also be true for $n = k+1$, and since we've verified the formula for the base case $n = 1$, by the principle of mathematical induction, the formula: $1^2 + 2^2 + 3^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6}$ is true for all positive integers $n$.

Summary:

• Base Case: Verified the formula works for $n = 1$.
• Inductive Hypothesis: Assumed the formula works for $n = k$.
• Inductive Step: Proved that if the formula works for $n = k$, it also works for $n = k+1$.
• Conclusion: The formula is true for all $n \geq 1$.

Would you like to explore any specific part of this proof further? Here are some questions that may help you deepen your understanding:

1. Why is it necessary to factor out $(k+1)$ in the inductive step?
2. What happens if the base case fails in an induction proof?
3. How can you apply mathematical induction to prove inequalities?
4. What are some examples of problems where induction fails or is not applicable?
5. How does the formula for the sum of squares compare with the sum of cubes formula?

Tip: When solving inductive proofs, carefully handling the algebra during the inductive step is crucial to ensuring a successful proof.