Let $n$ be a positive integer. Determine the first digit after the decimal point of the number

$$
\sum_{k=1}^{n} \frac{k(k+1)}{n}
$$
I. solution. It is known that the sum of the first $n$ positive integers is $\frac{n(n+1)}{2}$, and the sum of their squares is $\frac{n(n+1)(2 n+1)}{6}$. Using this, the sum can be transformed as follows:

$$
\begin{aligned}
\sum_{k=1}^{n} \frac{k(k+1)}{n} & =\frac{1}{n} \sum_{k=1}^{n} k^{2}+k=\frac{1}{n} \cdot\left(\frac{n(n+1)(2 n+1)}{6}+\frac{n(n+1)}{2}\right)= \\
& =\frac{(n+1)(2 n+1)}{6}+\frac{n+1}{2}=\frac{2 n^{2}+6 n+4}{6}= \\
& =\frac{n^{2}+3 n+2}{3}=\frac{(n+1)(n+2)}{3}
\end{aligned}
$$

The remainder of $n$ when divided by 3 can be 0, 1, or 2. If the remainder is 1 or 2, then one of $(n+1)$ and $(n+2)$ is divisible by 3, so the fraction is an integer, meaning there is a 0 after the decimal point. If $n$ is divisible by 3, then the remainder of $(n+1)$ is 1, and the remainder of $(n+2)$ is 2, and thus their product gives a remainder of 2. In this case, the first digit after the decimal point is 6.

II. solution. Using the fact that $\binom{n}{k}+\binom{n}{k+1}=\binom{n+1}{k+1}$, we perform the following transformations:

$$
\begin{aligned}
& \sum_{k=1}^{n} \frac{k(k+1)}{n}=\frac{2}{n} \cdot \sum_{k=1}^{n} \frac{k(k+1)}{2}=\frac{2}{n} \cdot \sum_{k=1}^{n}\binom{k+1}{2}= \\
& =\frac{2}{n} \cdot\left(\sum_{k=3}^{n}\binom{k+1}{2}+\binom{2}{2}+\binom{3}{2}\right)=\frac{2}{n} \cdot\left(\sum_{k=3}^{n}\binom{k+1}{2}+\binom{3}{3}+\binom{3}{2}\right)= \\
& =\frac{2}{n} \cdot\left(\sum_{k=4}^{n}\binom{k+1}{2}+\binom{4}{2}+\binom{4}{3}\right)=\frac{2}{n} \cdot\left(\sum_{k=5}^{n}\binom{k+1}{2}+\binom{5}{2}+\binom{5}{3}\right)= \\
& =\ldots=\frac{2}{n}\left(\binom{n+1}{2}+\binom{n+1}{3}\right)=\frac{2}{n} \cdot\binom{n+2}{3}= \\
& =\frac{2 \cdot(n+2)!}{n \cdot 3!\cdot(n-1)!}=\frac{(n+1)(n+2)}{3}
\end{aligned}
$$

If $n$ is not divisible by 3, then the value of $\frac{(n+1)(n+2)}{3}$ is an integer, so there is no decimal point (or if there is, the digit following it is 0). If $3 \mid n$, then the remainder of $(n+1)(n+2)$ when divided by 3 is 2, so the first digit after the decimal point is 6.