2. Calculate the sum $\frac{1}{2[\sqrt{1}]+1}+\frac{1}{2[\sqrt{2}]+1}+\cdots+\frac{1}{2[\sqrt{99}]+1}$, where $[x]$ represents the integer part of the real number $x$.
Solution and grading:

We observe that the first 3 fractions are equal to $\frac{1}{3}$, the next 5 fractions are equal to $\frac{1}{5}$. In general, we can observe that $\frac{1}{2[\sqrt{k}]+1}=\frac{1}{2 n+1}$ for any $k$ for which $[\sqrt{k}]=n$, i.e., $n \leqslant \sqrt{k}<n+1$, which is equivalent to $n^{2} \leqslant k \leqslant n^{2}+2 n$. Therefore, the equality $\frac{1}{2[\sqrt{k}]+1}=\frac{1}{2 n+1}$ holds for $2 n+1$ numbers (namely $\left.n^{2}, n^{2}+1, \ldots, n^{2}+2 n\right)$. We can group the fractions corresponding to these $2 n+1$ numbers; the sum of the fractions in each group is thus 1. We will have 9 such groups, corresponding to the values $n=1,2, \ldots, 9$. In conclusion, we will have:

$\frac{1}{2[\sqrt{1}]+1}+\frac{1}{2[\sqrt{2}]+1}+\cdots+\frac{1}{2[\sqrt{99}]+1}=\underbrace{\frac{1}{2[\sqrt{1}]+1}+\frac{1}{2[\sqrt{2}]+1}+\frac{1}{2[\sqrt{3}]+1}}_{=1}+$ $+\underbrace{\frac{1}{2[\sqrt{4}]+1}+\frac{1}{2[\sqrt{5}]+1}+\frac{1}{2[\sqrt{6}]+1}+\frac{1}{2[\sqrt{7}]+1}+\frac{1}{2[\sqrt{8}]+1}}_{=1}+\ldots+$ $+\underbrace{\frac{1}{2[\sqrt{81}]+1}+\frac{1}{2[\sqrt{82}]+1}+\ldots+\frac{1}{2[\sqrt{99}]+1}}_{=1}=\underbrace{1+1+\ldots+1}_{\text {9 times }}=9 \ldots \ldots \ldots .7 p$

Note: (partial scores)

Any calculation procedure that allows the final result to be calculated should be graded based on the progress made, whether it is based on the grouping from the above solution or explicitly lists the terms of the sum one by one.

For calculating a few terms in the sequence, 1 point will be awarded.

For observing that the terms repeat and correctly intuiting the number of repetitions (without calculating the sum), 3 points will be awarded.