4. On the board, the numbers $2,3,4, \ldots ., n+1$ are written. Then the products of every two of them are written, followed by the products of every three of them, and the process continues until the product of all the numbers 2,3,4,...., $n+1$ is written. Calculate the sum of the reciprocals of all the numbers written on the board.
Solution. Let's consider the product $\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right) \cdots\left(1+\frac{1}{n+1}\right)$. If all the multiplications are carried out, the desired sum of the reciprocal values increased by 1 is obtained. According to this, the desired sum is equal to

$$
\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right) \cdots\left(1+\frac{1}{n+1}\right)-1=\frac{3}{2} \cdot \frac{4}{3} \cdots \cdots \frac{n+2}{n+1}-1=\frac{n+2}{2}-1=\frac{n}{2} .
$$

## 4th Year