Find the sum of the finite geometric sequence.
$$\sum _ { n = 1 } ^ { 6 } 2 \left( - \frac { 2 } { 3 } \right) ^ { n - 1 }$$
A) $- \frac { 463 } { 243 }$
B) $- \frac { 1024 } { 3 }$
C) 4256
D) $\frac { 55 } { 81 }$
E) $\frac { 266 } { 243 }$

Question

Quizplus · Accepted Answer

The sum of a finite geometric sequence can be found using the formula $S_n = \frac{a(1 - r^n)}{1 - r}$, where $S_n$ is the sum of the first $n$ terms, $a$ is the first term, $r$ is the common ratio, and $n$ is the number of terms. In this case, $a = 2$, $r = -\frac{2}{3}$, and $n = 6$. Plugging these values into the formula gives:$$S_6 = \frac{2(1 - (-\frac{2}{3})^6)}{1 - (-\frac{2}{3})} = \frac{2(1 - \frac{64}{729})}{1 + \frac{2}{3}} = \frac{2(\frac{729 - 64}{729})}{\frac{5}{3}} = \frac{2 \cdot \frac{665}{729}}{\frac{5}{3}} = \frac{2 \cdot 665}{729} \cdot \frac{3}{5} = \frac{1330}{729} \cdot \frac{3}{5} = \frac{3990}{3645} = \frac{266}{243}.$$Therefore, the sum of the given finite geometric sequence is $\frac{266}{243}$.

Find the Sum of the Finite Geometric Sequence ∑n=162(−23)n−1\sum _ { n = 1 } ^ { 6 } 2 \left( - \frac { 2 } { 3 } \right) ^ { n - 1 }n=1∑6​2(−32​)n−1

Find the Sum of the Finite Geometric Sequence $\sum _ { n = 1 } ^ { 6 } 2 \left( - \frac { 2 } { 3 } \right) ^ { n - 1 }$