Find the fifth partial sum S₅ of the given series. $\sum _ { n = 1 } ^ { \infty } \frac { ( - 1 ) ^ { n } } { 3 ^ { n } }$ A) $- \frac { 121 } { 243 }$ B) $- \frac { 40 } { 81 }$ C) $- \frac { 61 } { 243 }$ D) $- \frac { 20 } { 81 }$

Question

Quizplus · Accepted Answer

The series is a geometric series with the first term $a = -\frac{1}{3}$ (when $n=1$) and common ratio $r = -\frac{1}{3}$. The fifth partial sum of a geometric series can be found using the formula $S_n = a \frac{1 - r^n}{1 - r}$, where $n$ is the number of terms. Plugging in the values, we get $S_5 = -\frac{1}{3} \frac{1 - (-\frac{1}{3})^5}{1 - (-\frac{1}{3})} = -\frac{1}{3} \frac{1 - (-\frac{1}{243})}{1 + \frac{1}{3}} = -\frac{1}{3} \frac{\frac{244}{243}}{\frac{4}{3}} = -\frac{1}{3} \cdot \frac{244}{243} \cdot \frac{3}{4} = -\frac{244}{243} \cdot \frac{1}{4} = -\frac{61}{243}$.

Find the Fifth Partial Sum S5 of the Given Series ∑n=1∞(−1)n3n\sum _ { n = 1 } ^ { \infty } \frac { ( - 1 ) ^ { n } } { 3 ^ { n } }∑n=1∞​3n(−1)n​

Find the Fifth Partial Sum S₅ of the Given Series $\sum _ { n = 1 } ^ { \infty } \frac { ( - 1 ) ^ { n } } { 3 ^ { n } }$