Sequence $A$.

The sequence of numbers $x_{1}, x_{2}, \ldots$ is such that $x_{1}=1 / 2$ and $x_{k+1}=x_{k}^{2}+x_{k}$ for any natural number $k$.

Find the integer part of the sum $\frac{1}{x_{1}+1}+\frac{1}{x_{1}+1}+\ldots+\frac{1}{x_{100}+1}$.
Note that $\frac{1}{x_{k+1}}=\frac{1}{x_{k}}-\frac{1}{x_{k}+1}$, which means $\frac{1}{x_{k}+1}=\frac{1}{x_{k}}-\frac{1}{x_{k+1}}$. Therefore,

$\frac{1}{x_{1}+1}+\frac{1}{x_{2}+1}+\ldots+\frac{1}{x_{100}+1}=\frac{1}{x_{1}}-\frac{1}{x_{2}}+\frac{1}{x_{2}}-\frac{1}{x_{3}}+\ldots+\frac{1}{x_{100}}-\frac{1}{x_{101}}=\frac{1}{x_{1}}-\frac{1}{x_{101}}=2-\frac{1}{x_{101}}$. The sequence is increasing, so $x_{101}>x_{3}>1$, hence the integer part of the desired sum is 1.

## Answer

1.