Answer
See below.
Work Step by Step
1. For $n=2$, we have $LHS=(1-\frac{1}{2^2})=\frac{3}{4}$ and $RHS=\frac{2+1}{2(2)}=\frac{3}{4}=LHS$, thus P(2) is true.
2. Assume P(k) ($k\gt2$) is true, we have:
$(1-\frac{1}{2^2})(1-\frac{1}{3^2})\cdots(1-\frac{1}{k^2})=\frac{k+1}{2(k)}$
3. For $n=k+1$, we have:
$LHS=(1-\frac{1}{2^2})(1-\frac{1}{3^2})\cdots(1-\frac{1}{k^2})(1-\frac{1}{(k+1)^2})\\=\frac{k+1}{2(k)}(1-\frac{1}{(k+1)^2})
=\frac{k+1}{2(k)}(\frac{k^2+2k}{(k+1)^2})=\frac{k+2}{2(k+1)}=RHS$
4. Thus $P(k+1)$ is also true and we have proved the statement by mathematical induction.