Answer:
\(a_n=\dfrac{1}{3}\cdot (3)^{n-1}\)
C is correct.
Step-by-step explanation:
We need to choose correct model by the graph which passes through the points (2,1) (3,3) and (4,9)
Option 1: \(a_n=\dfrac{1}{3}\cdot (27)^{n-1}\)
Put n=2 and to get a₂=1
\(a_2=\dfrac{1}{3}\cdot (27)^{2-1}\)
\(a_2=9\)
\(9
eq 1\)
False
Option 2: \(a_n=27\cdot (\dfrac{1}{3})^{n-1}\)
Put n=2 and to get a₂=1
\(a_2=27\cdot (\dfrac{1}{3})^{2-1}\)
\(a_2=9\)
\(9
eq 1\)
False
Option 3: \(a_n=\dfrac{1}{3}\cdot (3)^{n-1}\)
Put n=2 and to get a₂=1
\(a_2=\dfrac{1}{3}\cdot (3)^{2-1}\)
\(a_2=1\)
\(1= 1\)
TRUE
Similarly, we will check (3,3) and (4,9)
and we will get true
Hence, The sequence is \(a_n=\dfrac{1}{3}\cdot (3)^{n-1}\)
