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:
- Calculate \( \left( -\frac{1}{3} \right)^{6} = \frac{1}{729} \).
- So, \(1 - \left( -\frac{1}{3} \right)^{6} = 1 - \frac{1}{729} = \frac{729 - 1}{729} = \frac{728}{729}\).
- 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:
\[ \boxed{-1.75} \]
1 answer