Answer
The sum, ${{S}_{n}}$, of the first n terms of the sequence described in Exercise 1 is given by the formula ${{S}_{n}}=\underline{\frac{{{a}_{1}}\left( 1-{{r}^{n}} \right)}{1-r}}$, where ${{a}_{1}}$ is the first term and r is the common ratio, $ r\ne 1$.
Work Step by Step
A geometric sequence is a sequence in which each term is obtained after the first by multiplying the preceding term by a fixed nonzero constant.
Take the sequence,
$1,2,4,8,16\ldots $
In the above sequence, the common ratio between two consecutive terms is constant.
For example,
$\begin{align}
& \frac{2}{1}=\frac{4}{2} \\
& =\frac{8}{4} \\
& =\frac{16}{8} \\
& =2
\end{align}$
Here, the common ratio is a fixed nonzero constant, in this case 2.
So, the $ n $ th term (general term) of a geometric sequence is expressed by the formula,
${{a}_{n}}={{a}_{1}}{{r}^{n-1}}$
Here ${{a}_{1}}$ is the first term and $ r $ is the common ratio of the sequence.
Also, the sum,${{S}_{n}}$ of the first n terms of a geometric sequence is given by the formula,
${{S}_{n}}=\frac{{{a}_{1}}\left( 1-{{r}^{n}} \right)}{1-r}$
Here ${{a}_{1}}$ is the first term and $ r $ is the common ratio of the sequence except $ r\ne 1$.
