5. On the board, the numbers $1, \frac{1}{2}, \frac{1}{3}, \ldots, \frac{1}{100}$ are written. It is allowed to erase any two numbers $a$ and $b$ and write $a+b+a b$ instead. After several such operations, only one number remains on the board. What can this number be?
Answer: $2^{99}(1+1)(1+1 / 2) \ldots(1+1 / 100)-1=2^{99} 101-1$. Solution: It is necessary to show the associativity of the operation for three arbitrary $a, b, c$, and then perform the calculation, noting that at each step, an expression of the form $2^{k}\left(1+a_{1}\right)\left(1+a_{2}\right) \cdot \ldots \cdot\left(1+a_{k-1}\right)-1$ is obtained. Criteria: $\mp$ - answer without proof of independence from the order of operations.