Find a formula for the nth partial sum of the series and use it to find the series' sum if the series converges.
-$$5 + 45 + 405 + \ldots + 5 \cdot 9 ^ { n - 1 } + \ldots$$
A) $\frac { 5 \left( 1 - 9 ^ { n } \right) } { 1 - 9 }$; series diverges
B) $\frac { 5 ( 1 - 9 n ) } { 1 - 9 } ; \frac { 5 } { 8 }$
C) $\frac { 5 \left( 1 + 9 ^ { n } \right) } { 1 + 9 } ; \frac { 1 } { 2 }$
D) $\frac { 5 \left( 1 - 9 ^ { n } \right) } { 9 - 1 }$; series diverges

Question

Quizplus · Accepted Answer

The given series is a geometric series with the first term $a = 5$ and common ratio $r = 9$. The formula for the nth partial sum of a geometric series is $S_n = \frac{a(1 - r^n)}{1 - r}$. Substituting $a = 5$ and $r = 9$ gives $S_n = \frac{5(1 - 9^n)}{1 - 9}$, which matches option A. Since $|r| > 1$, the series diverges.

Find a Formula for the Nth Partial Sum of the Series