Question

Find the value of the series given below.
$2 + \dfrac{5}{{2!.3}} + \dfrac{{5.7}}{{3!{{.3}^2}}} + \dfrac{{5.7.9}}{{4!{{.3}^3}}} + .......$

Answer 1

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.