Question

Question
Evaluate sigma-summation Underscript lower i equals 1 overscript 6 EndScripts 7 left parenthesis negative one-third right parenthesis superscript lower i baseline. Round your answer to the nearest hundredth.
(1 point)
Responses

negative 0.01
Image with alt text: negative 0.01

5.25
5.25

negative 1.75
Image with alt text: negative 1.75

10.50

Answer 1

To evaluate the summation \(\sum_{i=1}^{6} 7 \left( -\frac{1}{3} \right)^{i}\), we can first factor out the \(7\) since it is a constant multiplier:

\[
= 7 \sum_{i=1}^{6} \left( -\frac{1}{3} \right)^{i}
\]

Next, we will calculate the inner summation \(\sum_{i=1}^{6} \left( -\frac{1}{3} \right)^{i}\). This is a geometric series with the first term \(a = -\frac{1}{3}\) and a common ratio \(r = -\frac{1}{3}\). The number of terms \(n = 6\).

The formula for the sum of the first \(n\) terms of a geometric series is:

\[
S_n = a \frac{1 - r^n}{1 - r}
\]

Plugging in the values:

\[
S_6 = -\frac{1}{3} \frac{1 - \left( -\frac{1}{3} \right)^{6}}{1 - \left( -\frac{1}{3} \right)}
\]

Calculating each component:

1. Calculate \( \left( -\frac{1}{3} \right)^{6} = \frac{1}{729} \).
2. So, \(1 - \left( -\frac{1}{3} \right)^{6} = 1 - \frac{1}{729} = \frac{729 - 1}{729} = \frac{728}{729}\).
3. The denominator becomes \(1 - \left( -\frac{1}{3} \right) = 1 + \frac{1}{3} = \frac{4}{3}\).

Now substituting back into the formula:

\[
S_6 = -\frac{1}{3} \cdot \frac{728/729}{4/3} = -\frac{1}{3} \cdot \frac{728}{729} \cdot \frac{3}{4} = -\frac{728}{4 \cdot 729} = -\frac{182}{729}
\]

Finally, we substitute this back into the expression for the total sum:

\[
7 S_6 = 7 \left( -\frac{182}{729} \right) = -\frac{1274}{729}
\]

Now we need to calculate \(-\frac{1274}{729}\) and round it to the nearest hundredth:

Calculating the division:

\[
-\frac{1274}{729} \approx -1.747
\]

Rounding to the nearest hundredth gives:

\[
-1.75
\]

Thus, the answer is:

\[
\boxed{-1.75}
\]