Find the n th partial sum of the geometric sequence. $-64,16,-4,1,-\frac{1}{4}, \ldots, \quad n=6$
A) $-\frac{819}{16}$
B) $-\frac{3}{16}$
C) $\frac{1}{16}$
D) $\frac{3277}{16}$
E) $-\frac{51}{16}$

Question

Quizplus · Accepted Answer

The sequence is geometric with the first term $a = -64$ and common ratio $r = -\frac{1}{4}$. The formula for the n-th partial sum of a geometric sequence is $S_n = a \frac{1-r^n}{1-r}$. Substituting $a = -64$, $r = -\frac{1}{4}$, and $n = 6$, we get $S_6 = -64 \frac{1-(-\frac{1}{4})^6}{1+\frac{1}{4}}$. Calculating this gives $S_6 = -\frac{819}{16}$.

Find the N Th Partial Sum of the Geometric Sequence −64,16,−4,1,−14,…,n=6-64,16,-4,1,-\frac{1}{4}, \ldots, \quad n=6−64,16,−4,1,−41​,…,n=6

Find the N Th Partial Sum of the Geometric Sequence $-64,16,-4,1,-\frac{1}{4}, \ldots, \quad n=6$