Geometric sequence equations. Game on.

Following the pattern:
1st term, or $n = 1\textsf{:}$ $2 \cdot 4^{0}$
2nd term, or $n = 2\textsf{:}$ $2 \cdot 4^{1}$
3rd term, or $n = 3\textsf{:}$ $2 \cdot 4^{2}$
4th term, or $n = 4\textsf{:}$ $2 \cdot 4^{3}$
Since each term has an exponent that is $1$ less than the term number, the $n$th term is $a_n = 2 \cdot 4^{n-1}$.

Following the pattern:
1st term, or $n = 1\textsf{:}$ $8 \cdot 2^{0}$
2nd term, or $n = 2\textsf{:}$ $8 \cdot 2^{1}$
3rd term, or $n = 3\textsf{:}$ $8 \cdot 2^{2}$
4th term, or $n = 4\textsf{:}$ $8 \cdot 2^{3}$
Since each term has an exponent that is $1$ less than the term number, the $n$th term is $f(n) = 8 \cdot 2^{n-1}$.

Following the pattern:
1st term, or $n = 1\textsf{:}$ $6 \cdot 3^{0}$
2nd term, or $n = 2\textsf{:}$ $6 \cdot 3^{1}$
3rd term, or $n = 3\textsf{:}$ $6 \cdot 3^{2}$
4th term, or $n = 4\textsf{:}$ $6 \cdot 3^{3}$
Since each term has an exponent that is $1$ less than the term number, the $n$th term is $w = 6 \cdot 3^{n-1}$.

Following the pattern:
1st term, or $n = 1\textsf{:}$ $10 \cdot 2^{0}$
2nd term, or $n = 2\textsf{:}$ $10 \cdot 2^{1}$
3rd term, or $n = 3\textsf{:}$ $10 \cdot 2^{2}$
4th term, or $n = 4\textsf{:}$ $10 \cdot 2^{3}$
Since each term has an exponent that is $1$ less than the term number, the $n$th term is $a_n = 10 \cdot 2^{n-1}$.

Following the pattern:
1st term, or $n = 1\textsf{:}$ $4 \cdot 5^{0}$
2nd term, or $n = 2\textsf{:}$ $4 \cdot 5^{1}$
3rd term, or $n = 3\textsf{:}$ $4 \cdot 5^{2}$
4th term, or $n = 4\textsf{:}$ $4 \cdot 5^{3}$
Since each term has an exponent that is $1$ less than the term number, the $n$th term is $f(n) = 4 \cdot 5^{n-1}$.

Following the pattern:
1st term, or $n = 1\textsf{:}$ $7 \cdot 2^{0}$
2nd term, or $n = 2\textsf{:}$ $7 \cdot 2^{1}$
3rd term, or $n = 3\textsf{:}$ $7 \cdot 2^{2}$
4th term, or $n = 4\textsf{:}$ $7 \cdot 2^{3}$
Since each term has an exponent that is $1$ less than the term number, the $n$th term is $w = 7 \cdot 2^{n-1}$.

Following the pattern:
1st term, or $n = 1\textsf{:}$ $9 \cdot 3^{0}$
2nd term, or $n = 2\textsf{:}$ $9 \cdot 3^{1}$
3rd term, or $n = 3\textsf{:}$ $9 \cdot 3^{2}$
4th term, or $n = 4\textsf{:}$ $9 \cdot 3^{3}$
Since each term has an exponent that is $1$ less than the term number, the $n$th term is $a_n = 9 \cdot 3^{n-1}$.

Following the pattern:
1st term, or $n = 1\textsf{:}$ $3 \cdot \left(\dfrac{1}{2}\right)^{0}$
2nd term, or $n = 2\textsf{:}$ $3 \cdot \left(\dfrac{1}{2}\right)^{1}$
3rd term, or $n = 3\textsf{:}$ $3 \cdot \left(\dfrac{1}{2}\right)^{2}$
4th term, or $n = 4\textsf{:}$ $3 \cdot \left(\dfrac{1}{2}\right)^{3}$
Since each term has an exponent that is $1$ less than the term number, the $n$th term is $w = 3 \cdot \left(\dfrac{1}{2}\right)^{n-1}$.