## SUBJECT I

Determine the non-zero natural number n such that $\frac{2}{3 \cdot 5}+\frac{2}{5 \cdot 7}+\ldots . .+\frac{2}{(n-2) \cdot n}=\frac{332}{999}$.
## SUBJECT I

$\frac{5-3}{3 \cdot 5}+\frac{7-5}{5 \cdot 7}+\ldots \ldots . . \frac{n-(n-2)}{(n-2) \cdot n}=\frac{332}{999}$

$\frac{5}{3 \cdot 5}-\frac{3}{3 \cdot 5}+\frac{7}{5 \cdot 7}-\frac{5}{5 \cdot 7}+\ldots \ldots . .+\frac{n}{(n-2) \cdot n}-\frac{(n-2)}{(n-2) \cdot n}=\frac{332}{999}$ $2 p$

$\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\ldots \ldots \ldots \ldots \ldots \ldots \ldots+\frac{1}{n-2}-\frac{1}{n}=\frac{332}{999}$

$\frac{1}{3}-\frac{1}{n}=\frac{332}{999} \Rightarrow \frac{1}{3}-\frac{332}{999}=\frac{1}{n}$

$\frac{1}{999}=\frac{1}{n} \Rightarrow n=999$