Answer:
[tex]\sum^{\infty}_0 {(-3)^n}[/tex]
Step-by-step explanation:
Notice that the series: 1 - 3 + 9 - 27 +... clearly has powers of the factor 3 in its terms and it is also an alternate series (the terms alternate from positive to negative). The terms are positive for [tex]a_0, a_2, a_4, ...[/tex] (even terms) , while the odd terms [tex]a_1, a_3, a_5, ...[/tex] are negative. (so most likely there should be a factor (-1) in the common ratio.
We can then represent it with the following summation expression:
[tex]\sum^{\infty}_0 {(-3)^n}[/tex] given that each of its first four terms are:
[tex]a_0=(-3)^0=1\\a_1=(-3)^1=-3\\a_2=(-3)^2=(-3)*(-3)=9\\a_3=(-3)^3=(-3)*(-3)*(-3)=-27[/tex]
