81. Calculate the product:

$$
P=\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)\left(1-\frac{1}{16}\right) \cdots\left(1-\frac{1}{n^{2}}\right)
$$
$\triangle$ Let's simplify this expression:

$$
P=\frac{2^{2}-1}{2^{2}} \cdot \frac{3^{2}-1}{3^{2}} \cdot \frac{4^{2}-1}{4^{2}} \cdots \cdots \cdot \frac{n^{2}-1}{n^{2}}=\frac{1 \cdot 3}{2^{2}} \cdot \frac{2 \cdot 4}{3^{2}} \cdot \frac{3 \cdot 5}{4^{2}} \cdots \cdot \frac{(n-1)(n+1)}{n^{2}}
$$

The obtained product can be reduced by $2 \cdot 3^{2} \cdot 4^{2} \cdot 5^{2} \cdot \cdots \cdot(n-1)^{2} \cdot n$.

After this, we will have: $P=\frac{n+1}{2 n}$.

Answer: $\frac{n+1}{2 n}$.