Respuesta :
The Solution.
The series is a geometric series since the condition below holds:
[tex]\begin{gathered} r=\frac{T_2}{T_1}=\frac{T_3}{T_2} \\ \text{Where T}_1=first\text{ term=2} \\ T_2=\sec ond\text{ term = 2(}\frac{-1}{4}) \\ \\ T_3=third\text{ term = 2(}\frac{-1}{4})^2 \end{gathered}[/tex]So, the common ratio (r) is:
[tex]r=\frac{T_2}{T_1}=\frac{2(-\frac{1}{4})}{2}=-\frac{1}{4}[/tex]The sum of the series is given as below:
[tex]\begin{gathered} S_n=\frac{a(1-r^n)}{1-r}\text{ , r}<1 \\ \text{Where s}_n=\text{ s um of n terms of the series} \\ a\text{ =first term = 2} \\ r\text{ =common ratio = -}\frac{1}{4} \\ n=\text{ number of terms = 6} \end{gathered}[/tex]Substituting the above values into the stated formula for sum, we get
[tex]S_6=\frac{2\lbrack(1-(-\frac{1}{4})^6\rbrack^{}}{1-(-\frac{1}{4})}[/tex][tex]\begin{gathered} S_6=\frac{2(1-\frac{1}{4096})}{1+\frac{1}{4}} \\ \\ S_6=\frac{2\times\frac{4095}{4096}}{\frac{5}{4}} \end{gathered}[/tex][tex]undefined[/tex]