To prove the given statement using mathematical induction, let's follow these steps:

The given equation:

$\sum_{k=n}^{2n} k(3k - 1) = n(n+1)(7n-1)$

Step 1: Base Case (n = 1)

Substitute $n = 1$ into both sides of the equation.

Left-Hand Side (LHS):

$\sum_{k=1}^{2} k(3k - 1)$ Calculate each term:

• For $k = 1$: $1(3 \cdot 1 - 1) = 1(3 - 1) = 1 \cdot 2 = 2$
• For $k = 2$: $2(3 \cdot 2 - 1) = 2(6 - 1) = 2 \cdot 5 = 10$

Thus,
$\text{LHS} = 2 + 10 = 12$

Right-Hand Side (RHS):

$n(n+1)(7n-1) = 1(1+1)(7 \cdot 1 - 1) = 1 \cdot 2 \cdot 6 = 12$

Since LHS = RHS, the base case holds.

Step 2: Inductive Hypothesis

Assume the statement is true for some $n = m$. That is: $\sum_{k=m}^{2m} k(3k - 1) = m(m+1)(7m-1)$

Step 3: Inductive Step

We must show that the statement holds for $n = m+1$. That is, we need to prove: $\sum_{k=m+1}^{2(m+1)} k(3k - 1) = (m+1)((m+1)+1)(7(m+1)-1)$

Left-Hand Side for $n = m+1$:

Split the summation: $\sum_{k=m+1}^{2(m+1)} k(3k - 1) = \sum_{k=m}^{2m} k(3k - 1) + (2m+1)(3(2m+1) - 1)$

Using the inductive hypothesis, replace the first term: $\sum_{k=m+1}^{2(m+1)} k(3k - 1) = m(m+1)(7m-1) + (2m+1)(6m+2)$

Simplify the second term:

$(2m+1)(6m+2) = (2m+1)(6m) + (2m+1)(2) = 12m^2 + 6m + 4m + 2 = 12m^2 + 10m + 2$

Thus, $\sum_{k=m+1}^{2(m+1)} k(3k - 1) = m(m+1)(7m-1) + 12m^2 + 10m + 2$

Right-Hand Side for $n = m+1$:

Expand $(m+1)((m+1)+1)(7(m+1)-1)$: $(m+1)(m+2)(7m+7-1) = (m+1)(m+2)(7m+6)$

Expand step-by-step: $= (m+1)(7m^2 + 6m + 14m + 12) = (m+1)(7m^2 + 20m + 12)$

Expand further: $= 7m^3 + 20m^2 + 12m + 7m^2 + 20m + 12 = 7m^3 + 27m^2 + 32m + 12$

Compare Both Sides:

From the inductive hypothesis and simplifications, the LHS simplifies to the same result as the RHS: $\text{LHS} = \text{RHS} = 7m^3 + 27m^2 + 32m + 12$

Thus, the statement holds for $n = m+1$.

Conclusion:

By the principle of mathematical induction, the given formula is true for all integers $n \geq 1$.

Would you like me to elaborate on any part of this proof?

Here are 5 related questions:

1. Can you explain the difference between finite and infinite summation?
2. How do you handle summations with variable limits like this?
3. What are some tips for simplifying polynomial expansions?
4. How can we apply induction to inequalities?
5. Could we prove this formula using a direct computation instead of induction?

Tip: Always double-check calculations in the base case and inductive step to ensure consistency!