Question

What is the sum of the first five terms in this series?(picture of problem below)

Answer 1

Provided information:

We know the first 4th terms of the series:

`6,-(6)/(3),(6)/(9),-(6)/(27)`

We can express this series in summation notation as:

`\sum_{i\mathop{=}1}^n(6)/((-3)^(i-1))`

Therefore, the 5th term of the series is:

`(6)/((-3)^(5-1))=(6)/((-3)^4)=(6)/(81)`

Now, we have to add the first five terms:

`6+(-(6)/(3))+(6)/(9)+(-(6)/(27))+(6)/(81)`

The next step is to convert all fractions to have the same denominator, so:

`\begin{gathered} 1st:6*(81)/(81)=(486)/(81) \\ 2nd:-(6)/(3)*(27)/(27)=-(162)/(81) \\ 3rd:(6)/(9)*(9)/(9)=(54)/(81) \\ 4th:-(6)/(27)*(3)/(3)=-(18)/(81) \end{gathered}`

Now, they have the same denominator, it remains the same and we just need to add the numerators:

`(486-162+54-18+6)/(81)=(366)/(81)`

And now, let's simplify the fraction by dividing the numerator and denominator by 3:

`((366)/(3))/((81)/(3))=(122)/(27)`

The answer is A. 122/27