To answer this question we first need to notice that the sequence of dots increases by a factor of 3 in each drawing. This means that this is a geometric sequence with common ratio 3.
We know that the nth term of a geometric sequence is given by:
[tex]a_n=a_1r^{n-1}[/tex]
where a1 is the first term and r is the common ratio. In this case a1=3 and r=3, hence the nth term is:
[tex]a_n=3(3)^{n-1}[/tex]
Now, if we add the first 15 stages, this means that we are adding the first 15 terms of the sequence, then our sum will be of the form:
[tex]\begin{gathered} a_1+a_2+a_3+a_4+\cdot\cdot\cdot+a_{15}_{} \\ =3(3)^{1-1}+3(3)^{2-1}+3(3)^{3-1}+3(3)^{4-1}+\cdot\cdot\cdot+3(3)^{15-1} \\ =3(3)^0+3(3)^1+3(3)^2+3(3)^3+\cdot\cdot\cdot+3(3)^{14} \\ =3(3^0+3^1+3^2+3^3+\cdot\cdot\cdot+3^{14}) \\ =3(1^{}+3^1+3^2+3^3+\cdot\cdot\cdot+3^{14}) \end{gathered}[/tex]
Then we notice that option a is a correct expression for the sum of the first 15 stages.
We also know that the sum of the first nth terms of a geometric sequence is given by:
[tex]\frac{a_1(1-r^n)}{1-r}[/tex]
plugging the values we know, we have that:
[tex]3\frac{(1-3^{15})}{1-3}[/tex]
Therefore option d is also a correct expression for the sum of the first 15 stages.
