Provided information:
We know the first 4th terms of the series:
[tex]6,-\frac{6}{3},\frac{6}{9},-\frac{6}{27}[/tex]
We can express this series in summation notation as:
[tex]\sum_{i\mathop{=}1}^n\frac{6}{(-3)^{i-1}}[/tex]
Therefore, the 5th term of the series is:
[tex]\frac{6}{(-3)^{5-1}}=\frac{6}{(-3)^4}=\frac{6}{81}[/tex]
Now, we have to add the first five terms:
[tex]6+(-\frac{6}{3})+\frac{6}{9}+(-\frac{6}{27})+\frac{6}{81}[/tex]
The next step is to convert all fractions to have the same denominator, so:
[tex]\begin{gathered} 1st:6*\frac{81}{81}=\frac{486}{81} \\ 2nd:-\frac{6}{3}*\frac{27}{27}=-\frac{162}{81} \\ 3rd:\frac{6}{9}*\frac{9}{9}=\frac{54}{81} \\ 4th:-\frac{6}{27}*\frac{3}{3}=-\frac{18}{81} \end{gathered}[/tex]
Now, they have the same denominator, it remains the same and we just need to add the numerators:
[tex]\frac{486-162+54-18+6}{81}=\frac{366}{81}[/tex]
And now, let's simplify the fraction by dividing the numerator and denominator by 3:
[tex]\frac{\frac{366}{3}}{\frac{81}{3}}=\frac{122}{27}[/tex]
The answer is A. 122/27
