Answer:
[tex]a_n=\dfrac{2^{n-1}}{9+3(n-1)}, n\geq 1[/tex]
Step-by-step explanation:
The sequence is given as:
[tex]\dfrac{1}{9}, \dfrac{2}{12},\dfrac{2^2}{15},\dfrac{2^3}{18}, \dfrac{2^4}{21} \cdots[/tex]
The numerator is a power of n with the starting term being [tex]2^0[/tex]
The denominator is being increased by 3 with the starting term being 9.
Therefore, the nth term of the sequence is:
[tex]a_n=\dfrac{2^{n-1}}{9+3(n-1)}, n\geq 1[/tex]
