Question

1. \\(\sum_{i = 1}^{8} (-6)^{i - 1}\\)

Explanation:

The given sum is a geometric series. For a geometric series \(\sum_{i = 1}^{n} a r^{i - 1}\), the sum formula is \(S_{n}=\frac{a(1 - r^{n})}{1 - r}\) when \(r
eq1\). Here, \(a = 1\) (when \(i = 1\), \((-6)^{1 - 1}=(-6)^{0}=1\)), \(r=-6\), and \(n = 8\).

Step 1: Identify \(a\), \(r\), and \(n\)

We have \(a = 1\), \(r=-6\), and \(n = 8\) for the geometric series \(\sum_{i = 1}^{8}(-6)^{i - 1}\).

Step 2: Apply the geometric series sum formula

Using the formula \(S_{n}=\frac{a(1 - r^{n})}{1 - r}\), substitute \(a = 1\), \(r=-6\), and \(n = 8\):
\[

$$\begin{align*} S_{8}&=\frac{1\times(1-(-6)^{8})}{1-(-6)}\\ &=\frac{1 - 1679616}{1 + 6}\\ &=\frac{-1679615}{7}\\ &=-239945 \end{align*}$$

\]

Answer:

\(-239945\)