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\[2 + \dfrac{5}{{2!.3}} + \dfrac{{5.7}}{{3!{{.3}^2}}} + \dfrac{{5.7.9}}{{4!{{.3}^3}}} + .......\]

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Hint:- Use the expansion of ${(1 + x)^n}$, where n is negative.

As, we are given with the series

$ \Rightarrow y = 2 + \dfrac{5}{{2!.3}} + \dfrac{{5.7}}{{3!{{.3}^2}}} + \dfrac{{5.7.9}}{{4!{{.3}^3}}} + ......$ (1)

So, the series at equation 1 can be written as,

$ \Rightarrow y = 1 + 1 + \dfrac{5}{{2!.3}} + \dfrac{{5.7}}{{3!{{.3}^2}}} + \dfrac{{5.7.9}}{{4!{{.3}^3}}} + ......{\text{ }}$ (2)

Now, we know that when we had to find the value of any typical series,

Then we try to manipulate the series into the expansion of a known function.

So, we had to manipulate the series at equation 2.

Above series can be manipulated as,

$ \Rightarrow y = 1 + \dfrac{3}{2}.\dfrac{2}{3} + \dfrac{{\dfrac{3}{2}.\dfrac{5}{2}}}{{2!}}.\dfrac{{{2^2}}}{{{3^2}}} + \dfrac{{\dfrac{3}{2}.\dfrac{5}{2}.\dfrac{7}{2}}}{{3!}}.\dfrac{{{2^3}}}{{{3^3}}} + \dfrac{{\dfrac{3}{2}.\dfrac{5}{2}.\dfrac{7}{2}.\dfrac{9}{2}}}{{4!}}.\dfrac{{{2^4}}}{{{3^4}}} + ......$ (3)

Now, as we know that the expansion of ${\left( {1 + x} \right)^n}$, where n is negative is,

$ \Rightarrow {(1 + x)^n} = 1 + \dfrac{n}{{1!}}x + \dfrac{{n(n - 1)}}{{2!}}{x^2} + \dfrac{{n(n - 1)(n - 2)}}{{3!}}{x^3} + ....$

As, we can see clearly that in the expansion of ${(1 + x)^n}$,

$ \Rightarrow $If we put $x = \dfrac{{ - 2}}{3}$ and ${\text{ }}n = \dfrac{{ - 3}}{2}$. Then it becomes the series given in equation 3.

$ \Rightarrow $So, series given in equation 3 is the expansion of ${\left( {1 - \dfrac{2}{3}} \right)^{\dfrac{{ - 3}}{2}}} = {\left( {\dfrac{1}{3}} \right)^{\dfrac{{ - 3}}{2}}} = {\left( 3 \right)^{\dfrac{3}{2}}} = 3\sqrt 3 $.

$ \Rightarrow $So, $y = 3\sqrt 3 $

$ \Rightarrow $Hence, the value of the given series will be ${\text{3}}\sqrt 3 $.

Note:- Whenever we came up with this type of problem then try to

manipulate the series into the expansion of a known function and then we can

the series in terms of that function. As this will be the easiest and efficient way

to find the solution to the problem.

As, we are given with the series

$ \Rightarrow y = 2 + \dfrac{5}{{2!.3}} + \dfrac{{5.7}}{{3!{{.3}^2}}} + \dfrac{{5.7.9}}{{4!{{.3}^3}}} + ......$ (1)

So, the series at equation 1 can be written as,

$ \Rightarrow y = 1 + 1 + \dfrac{5}{{2!.3}} + \dfrac{{5.7}}{{3!{{.3}^2}}} + \dfrac{{5.7.9}}{{4!{{.3}^3}}} + ......{\text{ }}$ (2)

Now, we know that when we had to find the value of any typical series,

Then we try to manipulate the series into the expansion of a known function.

So, we had to manipulate the series at equation 2.

Above series can be manipulated as,

$ \Rightarrow y = 1 + \dfrac{3}{2}.\dfrac{2}{3} + \dfrac{{\dfrac{3}{2}.\dfrac{5}{2}}}{{2!}}.\dfrac{{{2^2}}}{{{3^2}}} + \dfrac{{\dfrac{3}{2}.\dfrac{5}{2}.\dfrac{7}{2}}}{{3!}}.\dfrac{{{2^3}}}{{{3^3}}} + \dfrac{{\dfrac{3}{2}.\dfrac{5}{2}.\dfrac{7}{2}.\dfrac{9}{2}}}{{4!}}.\dfrac{{{2^4}}}{{{3^4}}} + ......$ (3)

Now, as we know that the expansion of ${\left( {1 + x} \right)^n}$, where n is negative is,

$ \Rightarrow {(1 + x)^n} = 1 + \dfrac{n}{{1!}}x + \dfrac{{n(n - 1)}}{{2!}}{x^2} + \dfrac{{n(n - 1)(n - 2)}}{{3!}}{x^3} + ....$

As, we can see clearly that in the expansion of ${(1 + x)^n}$,

$ \Rightarrow $If we put $x = \dfrac{{ - 2}}{3}$ and ${\text{ }}n = \dfrac{{ - 3}}{2}$. Then it becomes the series given in equation 3.

$ \Rightarrow $So, series given in equation 3 is the expansion of ${\left( {1 - \dfrac{2}{3}} \right)^{\dfrac{{ - 3}}{2}}} = {\left( {\dfrac{1}{3}} \right)^{\dfrac{{ - 3}}{2}}} = {\left( 3 \right)^{\dfrac{3}{2}}} = 3\sqrt 3 $.

$ \Rightarrow $So, $y = 3\sqrt 3 $

$ \Rightarrow $Hence, the value of the given series will be ${\text{3}}\sqrt 3 $.

Note:- Whenever we came up with this type of problem then try to

manipulate the series into the expansion of a known function and then we can

the series in terms of that function. As this will be the easiest and efficient way

to find the solution to the problem.