What is the sum of the infinite geometric series? -3-3/2-3/4-3/8-3/16

Accepted Solution

hmmm  [tex]\bf -3~,~-\cfrac{3}{2}~,~-\cfrac{3}{4}~,~-\cfrac{3}{8}~,~-\cfrac{3}{16}[/tex]

now, if you notice the terms there.... since we know is a geometric sequence/serie, we can tell there's a "common multiplier" namely a "common ratio", and if you divide any two terms, the latter by the former, the quotient you get is, yes, you guessed it, is the "common ratio", let's check about.

[tex]\bf -\cfrac{3}{2}\div -3\implies \cfrac{1}{2}\qquad \qquad -\cfrac{3}{4}\div -\cfrac{3}{2}\implies \cfrac{1}{2}[/tex]

so, as you can see, the "common ratio" is then, just 1/2.

now, for a geometric serie, whose common ratio is a fraction, namely less than 1 and greater than 0, then

[tex]\bf \textit{sum of an infinite geometric serie}\\\\ S=\sum\limits_{i=0}^{\infty}~a_1\cdot r^i\implies \cfrac{a_1}{1-r}\quad \begin{cases} a_1=\textit{first term's value}\\ r=\textit{common ratio}\\ \qquad 0\ \textless \ |r|\ \textless \ 1\\ ----------\\ a_1=-3\\ r=\frac{1}{2} \end{cases} \\\\\\ S=\cfrac{-3}{1-\frac{1}{2}}\implies S=\cfrac{-3}{\frac{1}{2}}\implies S=-6[/tex]