
What is the sum of the infinite geometric series $1 + \dfrac{1}{5} + \dfrac{1}{{25}} + .......$?
Answer
505.2k+ views
Hint: A series of numbers or quantities in geometric progression. In a geometric series the ratio of each two consecutive terms is a constant function. To check a given sequence is geometric check success entries in the sequence all have the same ratio.
Formula used:
For the infinite geometric series
$s = a + ar + a{r^2} + a{r^3} + .........$
$s = \dfrac{a}{{1 - ar}}$
$r = \dfrac{{\text{2nd term}}}{{\text{1st term}}} = \dfrac{{\text{3rd term}}}{{\text{2nd term}}}$
Complete step-by-step solution:
The given infinite geometric series is,
$1 + \dfrac{1}{5} + \dfrac{1}{{25}} + ...............$
There are infinite terms in the geometric series.
We have to use the formula of geometric series.
$s = \dfrac{a}{{1 - r}}$ As long as $\left| r \right| < 1$
Where,
$a$ is the first term
$r$ is the ratio between each term
There are infinity terms.
Hence first term $a = 1$
Ratio between the two terms is
$ = \dfrac{{\dfrac{1}{5}}}{1} = \dfrac{{\dfrac{1}{{25}}}}{{\dfrac{1}{5}}}$
\[ = \dfrac{1}{5}\]
Therefore ratio \[ = \dfrac{1}{5}\]\[ < 1\] so
Substitute the values in the equation\[s = \dfrac{a}{{1 - r}}\]
\[
a = 1 \\
r = \dfrac{1}{5} \\
\]
Apply the values,
\[ = \dfrac{1}{{1 - \dfrac{1}{5}}}\]
\[
= \dfrac{1}{{\dfrac{{5 - 1}}{5}}} \\
= \dfrac{1}{{\dfrac{4}{5}}} \\
\]
Multiple the values by \[5\] in numerator and denominator so we get,
\[s = \dfrac{1}{{\dfrac{4}{5}}} \times \dfrac{5}{5}\]
S\[ = \dfrac{5}{4}\]
Therefore the sum of the infinite geometric sequence is \[ = \dfrac{5}{4}\],
Note: A geometric series is a series for which the ratio of each two consecutive terms is a constant function of the summation index. The more general case of the ratio is a rational function of the summation index. Produces a series called a hyper geometric series. A geometric sequence is a sequence where the ratio or between terms is constant. A geometric series is the sum of the terms of geometric sequence. The nth partial sum of a geometric sequence can be calculated using the first term and common ratio. A geometric series is an infinite series whose terms are in a geometric progression or whose successive terms have a common ratio.
Formula used:
For the infinite geometric series
$s = a + ar + a{r^2} + a{r^3} + .........$
$s = \dfrac{a}{{1 - ar}}$
$r = \dfrac{{\text{2nd term}}}{{\text{1st term}}} = \dfrac{{\text{3rd term}}}{{\text{2nd term}}}$
Complete step-by-step solution:
The given infinite geometric series is,
$1 + \dfrac{1}{5} + \dfrac{1}{{25}} + ...............$
There are infinite terms in the geometric series.
We have to use the formula of geometric series.
$s = \dfrac{a}{{1 - r}}$ As long as $\left| r \right| < 1$
Where,
$a$ is the first term
$r$ is the ratio between each term
There are infinity terms.
Hence first term $a = 1$
Ratio between the two terms is
$ = \dfrac{{\dfrac{1}{5}}}{1} = \dfrac{{\dfrac{1}{{25}}}}{{\dfrac{1}{5}}}$
\[ = \dfrac{1}{5}\]
Therefore ratio \[ = \dfrac{1}{5}\]\[ < 1\] so
Substitute the values in the equation\[s = \dfrac{a}{{1 - r}}\]
\[
a = 1 \\
r = \dfrac{1}{5} \\
\]
Apply the values,
\[ = \dfrac{1}{{1 - \dfrac{1}{5}}}\]
\[
= \dfrac{1}{{\dfrac{{5 - 1}}{5}}} \\
= \dfrac{1}{{\dfrac{4}{5}}} \\
\]
Multiple the values by \[5\] in numerator and denominator so we get,
\[s = \dfrac{1}{{\dfrac{4}{5}}} \times \dfrac{5}{5}\]
S\[ = \dfrac{5}{4}\]
Therefore the sum of the infinite geometric sequence is \[ = \dfrac{5}{4}\],
Note: A geometric series is a series for which the ratio of each two consecutive terms is a constant function of the summation index. The more general case of the ratio is a rational function of the summation index. Produces a series called a hyper geometric series. A geometric sequence is a sequence where the ratio or between terms is constant. A geometric series is the sum of the terms of geometric sequence. The nth partial sum of a geometric sequence can be calculated using the first term and common ratio. A geometric series is an infinite series whose terms are in a geometric progression or whose successive terms have a common ratio.
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