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}
The explicit formula for a geometric sequence is:
u_{n}=u_{1}\cdot \left(q\right)^{n-1}
Where u_{1} is the first term of the sequence and q is the common ratio.
In our problem:
u_{1}=4
From the given recursive formula:
u_{n+1}=3\cdot u_{n}
The common ratio is 3.
The explicit formula for the given sequence is:
u_{n}=4\cdot\left(3\right)^{n-1}
\begin{cases} u_1=5 \cr \cr u_{n+1}=\dfrac{2}{3}\cdot u_n \text{ for } n≥1 \end{cases}
The explicit formula for a geometric sequence is:
u_{n}=u_{1}\cdot \left(q\right)^{n-1}
Where u_{1} is the first term of the sequence and q is the common ratio.
In our problem:
u_{1}=5
From the given recursive formula:
u_{n+1}=\dfrac{2}{3}\cdot u_{n}
The common ratio is \dfrac{2}{3}.
The explicit formula for the given sequence is:
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}
The explicit formula for a geometric sequence is:
u_{n}=u_{1}\cdot \left(q\right)^{n-1}
Where u_{1} is the first term of the sequence and q is the common ratio.
In our problem:
u_{1}=3
From the given recursive formula:
u_{n+1}=-2\cdot u_{n}
The common ratio is -2.
The explicit formula for the given sequence is:
u_{n}=3\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}
The explicit formula for a geometric sequence is:
u_{n}=u_{1}\cdot \left(q\right)^{n-1}
Where u_{1} is the first term of the sequence and q is the common ratio.
In our problem:
u_{1}=10
From the given recursive formula:
u_{n+1}=2\cdot u_{n}
The common ratio is 2.
The explicit formula for the given sequence is:
u_{n}=10\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}
The explicit formula for a geometric sequence is:
u_{n}=u_{1}\cdot \left(q\right)^{n-1}
Where u_{1} is the first term of the sequence and q is the common ratio.
In our problem:
u_{1}=8
From the given recursive formula:
u_{n+1}=\dfrac{1}{2}\cdot u_{n}
The common ratio is \dfrac{1}{2}.
The explicit formula for the given sequence is:
u_{n}=8\cdot\left( \dfrac{1}{2} \right)^{n-1}
\begin{cases} u_1=7 \cr \cr u_{n+1}=-3\cdot u_n \text{ for } n≥1 \end{cases}
The explicit formula for a geometric sequence is:
u_{n}=u_{1}\cdot \left(q\right)^{n-1}
Where u_{1} is the first term of the sequence and q is the common ratio.
In our problem:
u_{1}=7
From the given recursive formula:
u_{n+1}=-3\cdot u_{n}
The common ratio is -3.
The explicit formula for the given sequence is:
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}
The explicit formula for a geometric sequence is:
u_{n}=u_{1}\cdot \left(q\right)^{n-1}
Where u_{1} is the first term of the sequence and q is the common ratio.
In our problem:
u_{1}=2
From the given recursive formula:
u_{n+1}=2\cdot u_{n}
The common ratio is 2.
The explicit formula for the given sequence is:
u_{n}=\left(2\right)^{n}