2. Find the 1981st term of the sequence

$$
1,2,2,3,3,3,4,4,4,4,5,5,5,5,5, \ldots \ldots
$$
Solution. Let the 1981st term of the sequence be the number $k$. Note that the last occurrence of the number: 2 has the ordinal number 3, 3 has the ordinal number 6, 4 has the ordinal number 10, etc.; the last occurrence of the number $k$ in the sequence has the ordinal number $1+2+3+\ldots+k=\frac{k(k+1)}{2}$.

According to this, $\frac{k(k+1)}{2} \geq 1981$, from which we get that $k^{2}+k-3962 \geq 0$, i.e.

$$
k \in\left(-\infty, \frac{1}{2}(-1-\sqrt{15849})\right] \cup\left[\frac{1}{2}(-1+\sqrt{15849}),+\infty\right)
$$

Since $k>0$, it follows that $k$ is the smallest number that belongs to the interval $\left[\frac{1}{2}(-1+\sqrt{15849}),+\infty\right)$. We have: $\frac{1}{2}(-1+\sqrt{15849}) \approx 62.4$, so according to this, $k=63$.

Therefore, the 1981st term in the given sequence is the number 63.