To prove the given statement using Mathematical Induction, we need to prove that:

$\prod_{k=2}^{n} \left( 1 - \frac{1}{k^2} \right) = \frac{n+1}{2n}$

for all $n \in \mathbb{N}$ such that $n \geq 2$.

Step 1: Base Case ($n = 2$)

We begin by checking the case when $n = 2$.

The left-hand side of the equation when $n = 2$ is:

$\prod_{k=2}^{2} \left( 1 - \frac{1}{k^2} \right) = \left( 1 - \frac{1}{2^2} \right) = 1 - \frac{1}{4} = \frac{3}{4}$

Now, check the right-hand side for $n = 2$:

$\frac{2+1}{2 \times 2} = \frac{3}{4}$

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

Step 2: Inductive Step

Assume that the formula holds for some $n = k$, where $k \geq 2$. That is, we assume the following inductive hypothesis:

$\prod_{k=2}^{k} \left( 1 - \frac{1}{k^2} \right) = \frac{k+1}{2k}$

We need to show that the formula holds for $n = k+1$. That is, we need to prove that:

$\prod_{k=2}^{k+1} \left( 1 - \frac{1}{k^2} \right) = \frac{k+2}{2(k+1)}$

Step 3: Expand the Left-Hand Side for $n = k + 1$

We start with the left-hand side for $n = k + 1$:

$\prod_{k=2}^{k+1} \left( 1 - \frac{1}{k^2} \right) = \prod_{k=2}^{k} \left( 1 - \frac{1}{k^2} \right) \times \left( 1 - \frac{1}{(k+1)^2} \right)$

By the inductive hypothesis, we know:

$\prod_{k=2}^{k} \left( 1 - \frac{1}{k^2} \right) = \frac{k+1}{2k}$

Now, we multiply this by the next term $\left( 1 - \frac{1}{(k+1)^2} \right)$:

$\frac{k+1}{2k} \times \left( 1 - \frac{1}{(k+1)^2} \right)$

Simplify the second factor:

$1 - \frac{1}{(k+1)^2} = \frac{(k+1)^2 - 1}{(k+1)^2} = \frac{k(k+2)}{(k+1)^2}$

Thus, the product becomes:

$\frac{k+1}{2k} \times \frac{k(k+2)}{(k+1)^2}$

Simplify the expression:

$= \frac{k+1}{2k} \times \frac{k(k+2)}{(k+1)^2} = \frac{k(k+2)}{2k(k+1)} = \frac{k+2}{2(k+1)}$

This matches the right-hand side of the formula for $n = k+1$.

Step 4: Conclusion

Since both the base case and the inductive step hold true, by the principle of mathematical induction, the formula is valid for all $n \geq 2$:

$\prod_{k=2}^{n} \left( 1 - \frac{1}{k^2} \right) = \frac{n+1}{2n}$

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Would you like further details on any step or have more questions about mathematical induction?

Here are 5 related questions you could explore:

1. What are some common techniques for simplifying products like these?
2. How does the formula for a product of terms relate to telescoping series?
3. Can you prove the formula using a different method, like direct manipulation?
4. What are some other applications of mathematical induction in sequences or series?
5. How would you prove similar identities involving sums of fractions?

Tip: When working with products in induction, it often helps to break them into simpler components and simplify step-by-step to match the desired form.