1. **State the problem:** We are given a sequence $a_1=1$, $a_2=2$, $a_3=6$, $a_4=10$, $a_5=15$, and we want to find a formula for the $n$-th term $a_n$. 2. **Analyze the sequence:** Let's look at the differences between terms: $$a_2 - a_1 = 2 - 1 = 1$$ $$a_3 - a_2 = 6 - 2 = 4$$ $$a_4 - a_3 = 10 - 6 = 4$$ $$a_5 - a_4 = 15 - 10 = 5$$ The differences are not constant, so it's not arithmetic. Let's check if the sequence matches a known pattern. 3. **Check for triangular numbers:** The triangular number formula is: $$T_n = \frac{n(n+1)}{2}$$ Let's compute $T_n$ for $n=1$ to $5$: $$T_1 = \frac{1 \times 2}{2} = 1$$ $$T_2 = \frac{2 \times 3}{2} = 3$$ $$T_3 = \frac{3 \times 4}{2} = 6$$ $$T_4 = \frac{4 \times 5}{2} = 10$$ $$T_5 = \frac{5 \times 6}{2} = 15$$ 4. **Compare with given sequence:** The given sequence is $1, 2, 6, 10, 15$ but the triangular numbers are $1, 3, 6, 10, 15$. The only difference is at $a_2$ where the given term is 2 instead of 3. 5. **Hypothesis:** The sequence might be $a_n = T_n$ except for $n=2$. 6. **Check if the sequence can be expressed as $a_n = \frac{n(n+1)}{2} - f(n)$ where $f(n)$ is some correction term:** For $n=2$, $a_2 = 2$ but $T_2=3$, so $f(2) = 1$. For other $n$, $a_n = T_n$, so $f(n) = 0$ for $n \neq 2$. 7. **Conclusion:** The sequence matches triangular numbers except for the second term. If the problem intends the sequence to be triangular numbers, the formula is: $$a_n = \frac{n(n+1)}{2}$$ Otherwise, if the sequence is exactly as given, no simple closed formula fits all terms except a piecewise definition. **Final answer:** $$\boxed{a_n = \frac{n(n+1)}{2}}$$ This formula matches all terms except $a_2$ in the given sequence, which might be a typo or special case.