For a positive integer $n$, let $t_{n}=\frac{n(n+1)}{2}$. Writing down the last digits of $t_{1}=1, t_{2}=3, t_{3}=6, t_{4}=10, t_{5}=15 \cdots \cdots$ can form an infinite repeating decimal: $0.13605 \cdots$. Find the length of the repeating cycle of this decimal.
$$
\begin{array}{l}
t_{n+20}-t_{n}=\frac{(n+20)(n+20+1)}{2}-\frac{n(n+1)}{2} \\
=20 n+210=10(2 n+21)
\end{array}
$$

That is, the last digit of $t_{n+20}$ is the same as that of $t_{n}$.
Therefore, 20 is the length of the repeating cycle of this repeating decimal. This repeating decimal is
$$
0.13605186556815063100 .
$$