Question

How do you find ${{S}_{n}}$ for the geometric series ${{a}_{1}}=4,r=0.5,n=8$ ?

Answer 1

Hint: We have been given a geometric sequence whose first term is equal to 4, the fixed term with which the subsequent terms are to be multiplied is 0.5. We have to find the sum of the terms of this geometric sequence upto its eighth term. We shall find this geometric series by using the formula ${{S}_{n}}=\dfrac{a\left( 1-{{r}^{n}} \right)}{1-r}$.

Complete step by step solution:
Given the geometric series ${{a}_{1}}=4,r=0.5,n=8$.
We shall use the formula for calculating the sum of series of the given geometric sequence, ${{S}_{n}}=\dfrac{a\left( 1-{{r}^{n}} \right)}{1-r}$ where the symbols have their usual meanings as given in the problem because $r<1$.
In this formula, we shall put the values as $a=4,r=0.5$ and
$\Rightarrow {{S}_{n}}=\dfrac{4\left( 1-{{\left( 0.5 \right)}^{8}} \right)}{1-0.5}$
We know that $0.5=\dfrac{1}{2}$, thus modifying this value, we get
$\Rightarrow {{S}_{n}}=\dfrac{4\left( 1-{{\left( \dfrac{1}{2} \right)}^{8}} \right)}{1-\dfrac{1}{2}}$
$\Rightarrow {{S}_{n}}=\dfrac{4\left( 1-\dfrac{1}{{{2}^{8}}} \right)}{1-\dfrac{1}{2}}$
$\Rightarrow {{S}_{n}}=\dfrac{4\left( {{2}^{8}}-1 \right)}{{{2}^{8}}.\dfrac{1}{2}}$
We know that $4=2\times 2$. Thus, we can also write $4={{2}^{2}}$.
$\Rightarrow {{S}_{n}}=\dfrac{{{2}^{2}}\left( {{2}^{8}}-1 \right)}{{{2}^{8}}.\dfrac{1}{2}}$
Dividing ${{2}^{2}}$ from the numerator and denominator on the right hand side, we get
$\Rightarrow {{S}_{n}}=\dfrac{\left( {{2}^{8}}-1 \right)}{{{2}^{6}}.\dfrac{1}{2}}$
$\Rightarrow {{S}_{n}}=\dfrac{\left( {{2}^{8}}-1 \right)}{{{2}^{5}}}$
Here, ${{2}^{5}}$ represents multiplying 2 by itself 5 times and ${{2}^{8}}$ represents multiplying 2 by itself 8 times.
$\Rightarrow {{S}_{n}}=\dfrac{\left( 256-1 \right)}{32}$
$\Rightarrow {{S}_{n}}=\dfrac{255}{32}$
$\Rightarrow {{S}_{n}}=7.968$
Therefore, ${{S}_{n}}$ for the geometric series ${{a}_{1}}=4,r=0.5,n=8$ is equal to 7.968.

Note:
A sequence is an ordered list of numbers. In a geometric sequence, a term is the previous term times a fixed number (here this fixed number is equal to 0.5) but a geometric series is the sum of the terms in the geometric sequence. This sum can be calculated upto infinity by using the formula ${{S}_{\infty }}=\dfrac{a}{1-r}$. However, in this problem we were asked to calculate the geometric series only upto the eighth term of the geometric sequence.