We will prove the given statement by mathematical induction. The proposition we need to prove is:

$1 \times 4 + 2 \times 7 + 3 \times 10 + \cdots + n(3n+1) = n(n+1)^2$

Step 1: Base Case

We start by checking the case when $n = 1$.

For $n = 1$, the left-hand side (LHS) is: $1 \times (3(1) + 1) = 1 \times 4 = 4$

The right-hand side (RHS) is: $1(1+1)^2 = 1 \times 2^2 = 1 \times 4 = 4$

Since LHS = RHS, the base case holds true.

Step 2: Inductive Hypothesis

Now assume that the statement is true for some positive integer $k$, i.e.,

$1 \times 4 + 2 \times 7 + 3 \times 10 + \cdots + k(3k + 1) = k(k + 1)^2$

Step 3: Inductive Step

We need to prove that the statement also holds for $n = k + 1$, i.e.,

$1 \times 4 + 2 \times 7 + 3 \times 10 + \cdots + (k+1)(3(k+1) + 1) = (k+1)((k+1) + 1)^2$

Using the inductive hypothesis, the left-hand side for $n = k+1$ becomes:

$\left( 1 \times 4 + 2 \times 7 + \cdots + k(3k + 1) \right) + (k+1)(3(k+1) + 1)$

By the inductive hypothesis, we know that:

$1 \times 4 + 2 \times 7 + \cdots + k(3k + 1) = k(k+1)^2$

Thus, the left-hand side becomes:

$k(k+1)^2 + (k+1)(3(k+1) + 1)$

Simplify $3(k+1) + 1$: $3(k+1) + 1 = 3k + 3 + 1 = 3k + 4$

Now, the expression becomes:

$k(k+1)^2 + (k+1)(3k + 4)$

Factor out $(k+1)$ from both terms:

$(k+1) \left( k(k+1) + (3k+4) \right)$

Simplify the expression inside the parentheses:

$k(k+1) + (3k + 4) = k^2 + k + 3k + 4 = k^2 + 4k + 4$

Thus, we have:

$(k+1)(k^2 + 4k + 4) = (k+1)(k+2)^2$

This is exactly the right-hand side for $n = k+1$, i.e.,

$(k+1)((k+1)+1)^2$

Conclusion:

Since the statement holds for $n = 1$ and we have shown that if it holds for $n = k$, it also holds for $n = k+1$, by the principle of mathematical induction, the statement is true for all positive integers $n$.

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Do you want more details or have any questions?

Here are 5 questions to expand on this topic:

1. How does mathematical induction compare to other proof techniques?
2. Can you use mathematical induction for inequalities?
3. What are the common pitfalls in mathematical induction?
4. How can the inductive step be made simpler in some problems?
5. What is the role of the base case in an induction proof?

Tip: Always verify that both the base case and the inductive step are valid when applying induction. Missing either one can invalidate the proof.