
Determine whether sum to infinity exist $ 4,\,\,\dfrac{4}{3},\,\,\dfrac{4}{9},\,\,\dfrac{4}{{27}},\,\,\dfrac{4}{{81}}...... $ .
Answer
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Hint: Clearly from given series we see that numerator of the series is $ 4 $ in each fraction but denominator of each fraction is a multiple of $ 3 $ even also we see that each next fraction is $ \dfrac{1}{3} $ times of previous. Hence, given series is infinite G.P. Therefore, its sum can be easily obtained by using infinite G.P. series formula.
Sum of infinity G.P. series: $ \dfrac{a}{{1 - r}}, $ where ‘a’ is first term and ‘r’ is common ratio of the given G.P. series.
Complete step-by-step answer:
Given series is
$ 4,\,\,\dfrac{4}{3},\,\,\dfrac{4}{9},\,\,\dfrac{4}{{27}},\,\,\dfrac{4}{{81}}...... $
Or we can write above series as:
$ \dfrac{4}{{{3^0}}},\,\,\dfrac{4}{3},\,\,\dfrac{4}{9},\,\,\dfrac{4}{{27}},\,\,\dfrac{4}{{81}}......\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ {\because {3^0} = 1} \right\} $
Writing denominators as powers of $ '3' $ in above series we have:
$ \dfrac{4}{{{3^0}}} + \dfrac{4}{{{3^1}}} + \dfrac{4}{{{3^2}}} + \dfrac{4}{{{3^3}}}..........\infty $
Clearly, the above series formed inside the bracket is an infinite G.P. series with first term $ a = \dfrac{4}{{{3^0}}} = 4 $ and common ratio $ r = \dfrac{1}{3} $ .
To find its sum. We use the formula of infinite G.P. series which is $ \dfrac{a}{{1 - r}}. $
Substituting values of ‘a’ and ‘r’ we have
$
{S_\infty } = \dfrac{4}{{1 - (1/3)}} \\
\Rightarrow {S_\infty } = \dfrac{4}{{\dfrac{{3 - 1}}{3}}} \\
\Rightarrow {S_\infty } = \dfrac{4}{{\dfrac{2}{3}}} \\
\Rightarrow {S_\infty } = \dfrac{{12}}{2} \\
\Rightarrow {S_\infty } = 6 \;
$
Hence, from above we see that infinite sum of the series $ 4,\,\,\dfrac{4}{3},\,\,\dfrac{4}{9},\,\,\dfrac{4}{{27}},\,\,\dfrac{4}{{81}}...... $ is $ 6. $
Note: To find the sum of the given series we first check that the given series is either A.P. , G.P, natural series, A.G. or infinite G.P. series. And then after that we select appropriate formulas to find the sum with respect to the given series and finally on substituting values in formula so selected to get the required solution of the problem.
Sum of infinity G.P. series: $ \dfrac{a}{{1 - r}}, $ where ‘a’ is first term and ‘r’ is common ratio of the given G.P. series.
Complete step-by-step answer:
Given series is
$ 4,\,\,\dfrac{4}{3},\,\,\dfrac{4}{9},\,\,\dfrac{4}{{27}},\,\,\dfrac{4}{{81}}...... $
Or we can write above series as:
$ \dfrac{4}{{{3^0}}},\,\,\dfrac{4}{3},\,\,\dfrac{4}{9},\,\,\dfrac{4}{{27}},\,\,\dfrac{4}{{81}}......\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left\{ {\because {3^0} = 1} \right\} $
Writing denominators as powers of $ '3' $ in above series we have:
$ \dfrac{4}{{{3^0}}} + \dfrac{4}{{{3^1}}} + \dfrac{4}{{{3^2}}} + \dfrac{4}{{{3^3}}}..........\infty $
Clearly, the above series formed inside the bracket is an infinite G.P. series with first term $ a = \dfrac{4}{{{3^0}}} = 4 $ and common ratio $ r = \dfrac{1}{3} $ .
To find its sum. We use the formula of infinite G.P. series which is $ \dfrac{a}{{1 - r}}. $
Substituting values of ‘a’ and ‘r’ we have
$
{S_\infty } = \dfrac{4}{{1 - (1/3)}} \\
\Rightarrow {S_\infty } = \dfrac{4}{{\dfrac{{3 - 1}}{3}}} \\
\Rightarrow {S_\infty } = \dfrac{4}{{\dfrac{2}{3}}} \\
\Rightarrow {S_\infty } = \dfrac{{12}}{2} \\
\Rightarrow {S_\infty } = 6 \;
$
Hence, from above we see that infinite sum of the series $ 4,\,\,\dfrac{4}{3},\,\,\dfrac{4}{9},\,\,\dfrac{4}{{27}},\,\,\dfrac{4}{{81}}...... $ is $ 6. $
Note: To find the sum of the given series we first check that the given series is either A.P. , G.P, natural series, A.G. or infinite G.P. series. And then after that we select appropriate formulas to find the sum with respect to the given series and finally on substituting values in formula so selected to get the required solution of the problem.
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