Convert a geometric sequence between recursive and explicit form Precalculus

Find the explicit term of the sequences defined as follows.

\begin{cases} u_1=4 \cr \cr u_{n+1}=3u_n \text{ for } n≥1 \end{cases}

u_{n}=4\cdot\left(3\right)^{n-1}

u_{n}=4\cdot\left(3\right)^{n}

u_{n}=\dfrac{4}{3}\cdot\left(3\right)^{n-1}

u_{n}=\dfrac{4}{9}\cdot\left(3\right)^{n}

\begin{cases} u_1=5 \cr \cr u_{n+1}=\dfrac{2}{3}\cdot u_n \text{ for } n≥1 \end{cases}

u_{n}=5\cdot\left( \dfrac{2}{3} \right)^{n}

u_{n}=\dfrac{15}{2}\cdot\left( \dfrac{2}{3} \right)^{n-1}

u_{n}=5+\dfrac{2}{3}n

u_{n}=5\cdot\left( \dfrac{2}{3} \right)^{n-1}

\begin{cases} u_1=3 \cr \cr u_{n+1}=-2\cdot u_n \text{ for } n≥1 \end{cases}

u_{n}=3\cdot\left( -2 \right)^{n}

u_{n}=3\cdot\left( -2 \right)^{n-1}

u_{n}=-\dfrac{3}{2}\cdot\left( 2 \right)^{n}

u_{n}=-\dfrac{3}{2}\cdot\left( -2\right)^{n-1}

\begin{cases} u_1=10 \cr \cr u_{n+1}=2\cdot u_n \text{ for } n≥1 \end{cases}

u_{n}=10\cdot\left( 2 \right)^{n}

u_{n}=5\cdot\left( 2 \right)^{n-1}

u_{n}=10\cdot\left( 2 \right)^{n-1}

u_{n}=5\cdot\left(- 2 \right)^{n+1}

\begin{cases} u_1=8 \cr \cr u_{n+1}=\dfrac{1}{2}\cdot u_n \text{ for } n≥1 \end{cases}

u_{n}=8\cdot\left( \dfrac{1}{2} \right)^{n-1}

u_{n}=4\cdot\left( \dfrac{1}{2} \right)^{n+1}

u_{n}=4\cdot\left( \dfrac{1}{2} \right)^{n}

u_{n}=8\cdot\left( \dfrac{1}{2} \right)^{n}

\begin{cases} u_1=7 \cr \cr u_{n+1}=-3\cdot u_n \text{ for } n≥1 \end{cases}

u_{n}=7-3n

u_{n}=-\dfrac{7}{3}\cdot\left( 3\right)^{n}

u_{n}=7\cdot\left( -3\right)^{n}

u_{n}=7\cdot\left( -3\right)^{n-1}

\begin{cases} u_1=2 \cr \cr u_{n+1}=2\cdot u_n \text{ for } n≥1 \end{cases}

u_{n}=\left(2\right)^{n+1}

u_{n}=\left(2\right)^{n}

u_{n}=\left(2\right)^{n-1}

u_{n}=2n