Question

What is the sum of the first eight terms of a geometric series whose first term is 3 and whose common ratio is 1/2?

Answer 1

There is a formula...
Sn = a1 (1 - r^n) / (1 - r)
a1 = first term = 3
n = number of terms = 8
r = common ratio = 1/2

now we sub
S(8) = 3(1 - 1/2^8) / (1 - 1/2)
S(8) = 3(1 - (1/2^8) / (1/2)
S(8) = 3(1 - 1/256) / (1/2)
S(8) = 3 (256/256 - 1/256) / (1/2)
S(8) = 3(255/256) / (1/2)
S(8) = (765/256) / (1/2)
S(8) = 765/256 * 2/1
S(8) = 1530/256
S(8) = 765/128 or 5 125/128 or 5.9765625

Answer 2

$\bf \qquad \qquad \textit{sum of a finite geometric sequence}\\\\ S_n=\sum\limits_{i=1}^{n}\ a_1\cdot r^{i-1}\implies S_n=a_1\left( \cfrac{1-r^n}{1-r} \right)\quad \begin{cases} n=n^{th}\ term\\ a_1=\textit{first term's value}\\ r=\textit{common ratio}\\ ----------\\ a_1=3\\ r=\frac{1}{2}\\ n=8 \end{cases}$

$\bf S_8=3\left( \cfrac{1-\left( \frac{1}{2} \right)^8}{1-\frac{1}{2}} \right)\implies S_8=3\left( \cfrac{1-\frac{1^8}{2^8}}{1-\frac{1}{2}} \right) \\\\\\ S_8=3\left( \cfrac{1-\frac{1}{256}}{\frac{1}{2}} \right)\implies S_8=3\left( \cfrac{\frac{255}{256}}{\frac{1}{2}} \right)\implies S_8=3\left( \cfrac{255}{256}\cdot \cfrac{2}{1} \right) \\\\\\ S_8=3\left( \cfrac{255}{128} \right)\implies S_8=\cfrac{765}{128}$