Hello there. To solve this question, we'll have to remember some properties about geometric series.
Given that we want the sum of
[tex]1-3+9-27...[/tex]
First, we find the general term of this series:
Notice they are all powers of 3, namely
[tex]\begin{gathered} 1=3^0 \\ 3=3^1 \\ 9=3^2 \\ 27=3^3 \\ \vdots \end{gathered}[/tex]
But this is an alternating series, hence the general term is given by:
[tex]a_n=\left(-3\right)^{n-1}[/tex]
Since we just want the sum of the first 9 terms of this geometric series, we apply the formula:
[tex]S_n=\frac{a_1\cdot\left(1-q^n\right?}{1-q}[/tex]
Where q is the ratio between two consecutive terms of the series.
We find q as follows:
[tex]q=\frac{a_2}{a_1}=\frac{\left(-3\right)^{2-1}}{\left(-3\right)^{1-1}}=\frac{-3}{1}=-3[/tex]
Then we plug n = 9 in the formula, such that:
[tex]S_9=\frac{1\cdot\left(1-\left(-3\right)^9\right?}{1-\left(-3\right)}=\frac{1-\left(-19683\right)}{1+3}=\frac{19684}{4}[/tex]
Simplify the fraction by a factor of 4
[tex]S_9=4921[/tex]
This is the sum of the nine first terms of this geometric series and it is the answer contained in the second option.
