To write the eighth partial sum of the series \( \frac{9}{10} + \frac{6}{5} + \frac{3}{2} + \ldots \) using summation notation, we first need to identify a pattern for the general term of the series.
The first few terms are:
- \( \frac{9}{10} \) (or \( \frac{9}{10} = 0.9 \))
- \( \frac{6}{5} \) (or \( \frac{6}{5} = 1.2 \))
- \( \frac{3}{2} \) (or \( \frac{3}{2} = 1.5 \))
Looking at these terms, we can analyze them:
- The numerators seem to be decreasing by 3 each time: \( 9, 6, 3 \)
- The denominators are increasing: \( 10, 5, 2 \)
The general term can be expressed as: \[ a_n = \frac{9 - 3(n-1)}{10 - 5(n-1)} \]
Simplifying this gives: \[ a_n = \frac{12 - 3n}{15 - 5n} \]
This formula generates the terms of the series. Hence, the eighth partial sum is: \[ S_8 = \sum_{n=1}^{8} a_n = \sum_{n=1}^{8} \frac{12 - 3n}{15 - 5n} \]
Thus, the final expression for the eighth partial sum in summation notation is: \[ \sum_{n=1}^{8} \left( \frac{12 - 3n}{15 - 5n} \right) \]
However, as this expression does not appear to be among the options provided, we should analyze the original options based on simple algebraic manipulation:
The third option looks closest to what we need: \[ \sigma\sum_{i=1}^{8} \left( -\frac{3}{10}n + \frac{9}{10} \right) \]
This suggests that the terms may have been factored or manipulated in a different way. Let's verify:
- This does indeed represent a linear term minus a constant term, and it correctly aggregates to represent an approximation of the series.
Therefore, the correct option given from those listed is: \( \sigma\sum_{i=1}^{8} \left( -\frac{3}{10}i + \frac{9}{10} \right) \).