Question

Find the sum of the first 8 terms of the series. 1/2 + 1 + 2 + 4

Answer 1

$\bf \cfrac{1}{2}~~,~~\stackrel{2\cdot \frac{1}{2}}{1}~~,~~\stackrel{2\cdot 1}{2}~~,~~\stackrel{2\cdot 2}{4}~~,...$

so as you can see the common ratio is 2, and the first term is 1/2,

$\bf \qquad \qquad \textit{sum of a finite geometric sequence} \\\\ S_n=\sum\limits_{i=1}^{n}\ a_1\cdot r^{i-1}\implies S_n=a_1\left( \cfrac{1-r^n}{1-r} \right)\quad \begin{cases} n=n^{th}\ term\\ a_1=\textit{first term's value}\\ r=\textit{common ratio}\\ ----------\\ r=2\\ a_1=\frac{1}{2}\\ n=8 \end{cases} \\\\\\ S_8=\cfrac{1}{2}\left( \cfrac{1-2^8}{1-2} \right)\implies S_8=\cfrac{1}{2}\left( \cfrac{-255}{-1} \right)\implies S_8=\cfrac{255}{2}$

Answer 2

Answer:

The sum of 8 term of the series is $\dfrac{255}{2}$ or 127.5

Step-by-step explanation:

Given: The series $\frac{1}{2}$ + 1 +2 +4 + ...

We need to find the sum of 8 term

Common ratio, r $=\dfrac{2}{1}= 2$

First term, $a=\dfrac{1}{2}$

n=8

Formula:

$S_n=\dfrac{a(r^n-1)}{(r-1)}$

Substitute the value of a, r and n into formula

Sum of 8th term of the sequence

$S_8=\dfrac{1/2(2^8-1)}{2-1}$

$S_8=\dfrac{255}{2}$

Hence, The sum of 8 term of the series is $\dfrac{255}{2}$ or 127.5