sum_(k=1)^(oo)(-1)^(k)(2^(k-1))/(5^(k))=

Recognize as Geometric Series: The given series is an infinite alternating series of the form sum_(k=1)^(oo)(-1)^(k)(2^(k-1))/(5^(k)). To find the sum of this series, we can recognize it as a geometric series with the first term a and common ratio r. The first term a is obtained when k=1, which gives us (-1)^(1)(2^(1-1))/(5^(1)) = -1/5. The common ratio r is the factor by which each term is multiplied to get the next term. To find r, we can take the ratio of the second term to the first term. The second term is (-1)^(2)(2^(2-1))/(5^(2)) = 2/25, and the ratio r is (2/25) / (-1/5) = -2/5. Find First Term and Common Ratio: The sum S of an infinite geometric series with first term a and common ratio r, where |r| < 1, is given by S = a / (1 - r). In our case, a = -1/5 and r = -2/5. We can now substitute these values into the formula to find the sum of the series. Calculate Sum of Series: Substituting the values of a and r into the formula, we get S = (-1/5) / (1 - (-2/5)) = (-1/5) / (1 + 2/5) = (-1/5) / (5/5 + 2/5) = (-1/5) / (7/5). To divide by a fraction, we multiply by its reciprocal, so S = (-1/5) * (5/7) = -1/7.

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sum_(k=1)^(oo)(-1)^(k)(2^(k-1))/(5^(k))=

$\sum_{k=1}^{\infty}(-1)^{k} \frac{2^{k-1}}{5^{k}}=$

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Algebra 2

Sum of finite series starts from 1

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\sum_{k=1}^{\infty}(-1)^{k} \frac{2^{k-1}}{5^{k}}=

Recognize as Geometric Series: The given series is an infinite alternating series of the form $\sum_{k=1}^{\infty}(-1)^{k}\frac{2^{k-1}}{5^{k}}$ . To find the sum of this series, we can recognize it as a geometric series with the first term $a$ and common ratio $r$ . The first term $a$ is obtained when $k=1$ , which gives us $(-1)^{1}\frac{2^{1-1}}{5^{1}} = -\frac{1}{5}$ . The common ratio $r$ is the factor by which each term is multiplied to get the next term. To find $r$ , we can take the ratio of the second term to the first term. The second term is $(-1)^{2}\frac{2^{2-1}}{5^{2}} = \frac{2}{25}$ , and the ratio $r$ is $a$ $0$ .
Find First Term and Common Ratio: The sum $S$ of an infinite geometric series with first term $a$ and common ratio $r$ , where |r| < 1, is given by $S = \frac{a}{1 - r}$ . In our case, $a = -\frac{1}{5}$ and $r = -\frac{2}{5}$ . We can now substitute these values into the formula to find the sum of the series.
Calculate Sum of Series: Substituting the values of $a$ and $r$ into the formula, we get $S = \frac{-1}{5} / \left(1 - \left(-\frac{2}{5}\right)\right) = \frac{-1}{5} / \left(1 + \frac{2}{5}\right) = \frac{-1}{5} / \left(\frac{5}{5} + \frac{2}{5}\right) = \frac{-1}{5} / \left(\frac{7}{5}\right)$ . To divide by a fraction, we multiply by its reciprocal, so $S = \frac{-1}{5} \cdot \frac{5}{7} = -\frac{1}{7}$ .