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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?

Respuesta :

[tex]\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}[/tex]

[tex]\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}[/tex]
abidemiokin

The sum of the first eight terms of a geometric series whose first term is 3 and whose common ratio is 1/2 is 5.9765625

Sum of geometric series

Series are the sum of sequences. The sum of geometric series is expressed as:

Sn = a(1-r^n)/1-r

Given the following

n = 8

a = 3

r = 1/2

Substitute

S8 = 3(1-(1/2)^8)/0.5
S8 = 3(1-1/256)/0.5
S8 = 3(255/256)/0.5
S8 = 5.9765625

Hence the sum of the first eight terms of a geometric series whose first term is 3 and whose common ratio is 1/2 is 5.9765625

Learn more on series here: https://brainly.com/question/723406

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