1. The problem is to write the sum $9 + 4 - 1 - ... - 16$ using sigma notation starting at index 1. 2. We are told the sequence is arithmetic with common difference $-5$. 3. The first term $a_1$ is 9. 4. The $n$th term of an arithmetic sequence is given by the formula: $$a_n = a_1 + (n-1)d$$ where $d$ is the common difference. 5. Substitute $a_1 = 9$ and $d = -5$: $$a_n = 9 + (n-1)(-5) = 9 - 5(n-1)$$ 6. Simplify the expression: $$a_n = 9 - 5n + 5 = 14 - 5n$$ 7. To find the number of terms, note the last term is $-16$: Set $a_n = -16$: $$14 - 5n = -16$$ $$-5n = -16 - 14 = -30$$ $$n = \frac{-30}{-5} = 6$$ 8. Therefore, the sum can be written as: $$\sum_{n=1}^6 (14 - 5n)$$ This sigma notation represents the sum $9 + 4 - 1 - 6 - 11 - 16$. Final answer: $$\sum_{n=1}^6 (14 - 5n)$$