For $n \geq 2$, prove that $(1- \frac{1}{4})(1- \frac{1}{9})(1- \frac{1}{16})...(1- \frac{1}{n^2}) = \frac{n+1}{2n}$ For $n \geq 2$ prove that $(1- \frac{1}{4})(1- \frac{1}{9})(1- \frac{1}{16})...(1- \frac{1}{n^2}) = \frac{n+1}{2n}$
We need to use induction. 
The Principle of Mathematical Induction, Theorem 4.2.1, states that $n_0 \in \mathbb{Z}$. For each integer $n \geq n_o$, let $P(n)$ be a statement about $n$. Suppose that the following two statements are true:


*

*$P(n_o)$

*$( \forall n \geq n_o)[P(n) \rightarrow P(n+1)]$


Then, for all integers $n \geq n_o$, that statement $P(n)$ is true.
Suppose, 
$(1- \frac{1}{4})(1- \frac{1}{9})(1- \frac{1}{16})...(1- \frac{1}{n^2}) = \frac{n+1}{2n}$, the equation, is true for some natural number $k$.
$\frac{1}{2^2})*...*(1- \frac{1}{k^2}) = \frac{k+1}{2k}$
The induction hypothesis $P(k) \rightarrow P(k+1)$ must hold for any natural number $k, P(k),$ from the basis and inductive step.
$(\frac{1}{2^2})*...*(1- \frac{1}{k^2})*(1- \frac{1}{(k+1)^2}) = \frac{k+1}{2k}(1- \frac{1}{(k+1)^2})$
$\leftrightarrow$ $\frac{k+1}{2k}(\frac{(k+1)^2}{(k+1)^2}- \frac{1}{(k+1)^2})$
$\leftrightarrow$ $(\frac{(k+1)}{2k})( \frac{(k+1)^2-1}{(k+1)^2})$
$\leftrightarrow$ $( \frac{(k^2+2k+1-1}{(2k)(k+1)})$
$\leftrightarrow$ $( \frac{(k^2+2k}{(2k)(k+1)})$
$\leftrightarrow$ $( \frac{k+2}{(2)(k+1)})$
We need to demonstrate that 
$(\frac{1}{2^2})*...*(1- \frac{1}{n^2}) = \frac{n+1}{2n}$
holds true for $n \geq 2$
The question is do I just substitute a number that is greater than two after I did the induction?

My way doesn't use induction. It uses formulas $a^2-b^2=(a-b)(a+b)$
$$(1- \frac{1}{4})(1- \frac{1}{9})(1- \frac{1}{16})...(1- \frac{1}{n^2}) =\\=((1- \frac{1}{2})(1- \frac{1}{3})(1- \frac{1}{4})...(1- \frac{1}{n}))((1+ \frac{1}{2})(1+ \frac{1}{3})(1+\frac{1}{4})...(1+ \frac{1}{n})) = \\=(\frac12\cdot\frac23\cdot\frac34 \ldots \cdot\frac{n-1}{n}) (\frac32\cdot\frac43\cdot\frac54 \ldots \cdot\frac{n+1}{n})= \frac{1}{n}(\frac{1}{2}(n+1))=\frac{n+1}{2n}$$
