Evaluate^6 \sum_(i=1)^6 (-(1)/(3))^(i). Round your answer to the nearest hundredth.
-0.01
5.25
-1.75
10.50
We can find the sum by using the formula for the sum of a geometric series:
\[ \sum_{i=1}^n ar^{i-1} = \frac{a(1-r^n)}{1-r} \]
where in this case, \( a = - \frac{1}{3} \) and \( r = - \frac{1}{3} \).
Plugging in the values, we have:
\[
\sum_{i=1}^6 \left(- \frac{1}{3}\right)^i = \frac{(-1/3)(1-(1/3)^6)}{1-(-1/3)} \approx - 1.75
\]
Rounded to the nearest hundredth, the answer is \boxed{-1.75}.
Evaluate \sum_(i=1)^6 (-(1)/(3))^(i). Round your answer to the nearest hundredth.
-0.01
5.25
-1.75
10.50
We can find the sum by using the formula for the sum of a geometric series:
\[ \sum_{i=1}^n ar^{i-1} = \frac{a(1-r^n)}{1-r} \]
where in this case, \( a = - \frac{1}{3} \) and \( r = - \frac{1}{3} \).
Plugging in the values, we have:
\[
\sum_{i=1}^6 \left(- \frac{1}{3}\right)^i = \frac{(-1/3)(1-(1/3)^6)}{1-(-1/3)} \approx -1.75
\]
Rounded to the nearest hundredth, the answer is \boxed{-1.75}.
show steps
To find the sum, let's begin by expanding the series and combining like terms. We have:
\[
\sum_{i=1}^6 \left(- \frac{1}{3}\right)^i = \left(- \frac{1}{3}\right)^1 + \left(- \frac{1}{3}\right)^2 + \left(- \frac{1}{3}\right)^3 + \left(- \frac{1}{3}\right)^4 + \left(- \frac{1}{3}\right)^5 + \left(- \frac{1}{3}\right)^6
\]
Simplifying the fractions, we have:
\[
-\frac{1}{3} - \frac{1}{9} - \frac{1}{27} - \frac{1}{81} - \frac{1}{243} - \frac{1}{729}
\]
To combine these fractions, we need to find a common denominator. In this case, the smallest common denominator is 729. So, we can rewrite the fractions with the common denominator:
\[
-\frac{243}{729} - \frac{81}{729} - \frac{27}{729} - \frac{9}{729} - \frac{3}{729} - \frac{1}{729}
\]
Now, we can combine the fractions:
\[
-\frac{243+81+27+9+3+1}{729} = -\frac{364}{729}
\]
Finally, let's round the fraction to the nearest hundredth:
\[
-\frac{364}{729} \approx -1.75
\]
Therefore, the sum is approximately -1.75.
Evaluate \sum_i=1^6 (-1/3)^i. Round your answer to the nearest hundredth.
-0.01
5.25
-1.75
10.50
To evaluate the sum, we can use the formula for the sum of a geometric series:
\[ \sum_{i=1}^n ar^{i-1} = \frac{a(1-r^n)}{1-r} \]
In this case, \( a = -\frac{1}{3} \) and \( r = -\frac{1}{3} \). Plugging in these values, we have:
\[ \sum_{i=1}^6 \left(-\frac{1}{3}\right)^i = \frac{-\frac{1}{3}(1-\left(-\frac{1}{3}\right)^6)}{1-\left(-\frac{1}{3}\right)} \]
Simplifying the numerator, we have:
\[ \frac{-\frac{1}{3}\left(1-\frac{1}{729}\right)}{1+\frac{1}{3}} \]
Further simplifying the numerator:
\[ \frac{-\frac{1}{3}\left(\frac{728}{729}\right)}{\frac{4}{3}} \]
Simplifying the denominator:
\[ \frac{-\frac{1}{3}\cdot\frac{728}{729}}{\frac{4}{3}} \]
Multiplying the fraction by its reciprocal:
\[ \frac{-728}{729}\cdot\frac{3}{4} \]
Cancelling out common factors:
\[ \frac{-364}{243} \]
Rounded to the nearest hundredth, the answer is approximately \boxed{-1.50}.