Question

Find the sum of the first 9 terms in the following geometric series.

Answer 1

Answer:

The sum of the first 9 terms is $S_9=\frac{511}{4}$.

Step-by-step explanation:

To find the sum of the first $S_n$ terms of a geometric sequence use the formula

                                                  $S_n=\frac{a_1(1-r^n)}{1-r}$

where, $n$ is the number of terms, $a_1$ is the first term and $r$ is the common ratio.

To find the common ratio, find the ratio between a term and the term preceding it.  

Given the geometric sequence $64+32+16+...$, the common ratio is

$r=\frac{32}{64} =\frac{1}{2}$

and the sum of the first 9 terms is

$S_9=\frac{64(1-(\frac{1}{2})^9)}{1-\frac{1}{2}}\\\\S_9=\frac{64\left(-\left(\frac{1}{2}\right)^9+1\right)}{\frac{1}{2}}\\\\S_9=\frac{64\left(-\frac{1}{2^9}+1\right)}{\frac{1}{2}}\\\\S_9=\frac{64\left(-\frac{1}{512}+1\right)}{\frac{1}{2}}\\\\S_9=\frac{64\left(1-\frac{1}{512}\right)\cdot \:2}{1}=\frac{128\left(-\frac{1}{512}+1\right)}{1}\\\\S_9=\frac{128\cdot \frac{511}{512}}{1}=\frac{511\cdot \:128}{512}=\frac{65408}{512}=\frac{511}{4}$

Answer 2

Answer: 127.75.

(Got it wrong, clicked on help, and it showed me the answer. Done.)