Answer: Choice D [tex]\textrm{v}_n = 15(3^n)[/tex]
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Work Shown:
[tex]\textrm{v}_1 = 45[/tex] is the first term
[tex]\textrm{v}_n = 3\textrm{v}_{n-1}[/tex] means we multiply the previous term ([tex]\textrm{v}_{n-1}[/tex]) by 3 to get the next term ([tex]\textrm{v}_n[/tex]). Therefore, r = 3 is the common ratio. This sequence is geometric.
Let's find the explicit formula
[tex]\textrm{v}_n = a*r^{n-1}[/tex]
[tex]\textrm{v}_n = \textrm{v}_1*r^{n-1}[/tex]
[tex]\textrm{v}_n = 45*3^{n-1}[/tex]
[tex]\textrm{v}_n = 45*3^n*3^{-1}[/tex]
[tex]\textrm{v}_n = 45*3^{-1}*3^n[/tex]
[tex]\textrm{v}_n = \left(45*\frac{1}{3}\right)*(3^n)[/tex]
[tex]\textrm{v}_n = 15(3^n)[/tex]
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For each sequence mentioned, the starting term is at n = 1.
So to check our work, we can plug n = 1 into the equation we just found to get...
[tex]\textrm{v}_n = 15*(3^n)[/tex]
[tex]\textrm{v}_1 = 15*(3^1)[/tex]
[tex]\textrm{v}_1 = 45[/tex]
The other terms are generated in a similar fashion.
