Question

Find the sum of the infinite series $\dfrac{3}{4} + \dfrac{{3.5}}{{4.8}} + \dfrac{{3.5.7}}{{4.8.12}} + ....$

Answer 1

Hint: Let $S = \dfrac{3}{4} + \dfrac{{3.5}}{{4.8}} + \dfrac{{3.5.7}}{{4.8.12}} + ....$. Simplify the equation by adding 1 on both sides. Use the expansion \[{(1 - x)^{ - \dfrac{p}{q}}} = 1 + \dfrac{p}{{1!}}(\dfrac{x}{q}) + \dfrac{{p(p + q)}}{{2!}}{(\dfrac{x}{q})^2} + ....\] and compare its RHS with that of S to get the value of x and hence find the value of S.

Complete step by step solution:
We have an infinite series $\dfrac{3}{4} + \dfrac{{3.5}}{{4.8}} + \dfrac{{3.5.7}}{{4.8.12}} + ....$
We need to find the sum of this series.
Let’s call the sum as S.
Then we have $S = \dfrac{3}{4} + \dfrac{{3.5}}{{4.8}} + \dfrac{{3.5.7}}{{4.8.12}} + ....(1)$
We will simplify the above expression by adding 1 on both the sides of the equation.
Thus, we have
$1 + S = 1 + \dfrac{3}{4} + \dfrac{{3.5}}{{4.8}} + \dfrac{{3.5.7}}{{4.8.12}} + ....$
Now, RHS can be expressed as follows:
$1 + S = 1 + \dfrac{3}{{1!}}.(\dfrac{1}{4}) + \dfrac{{3.5}}{{2!}}.{(\dfrac{1}{4})^2} + \dfrac{{3.5.7}}{{3!}}.{(\dfrac{1}{4})^3} + ....(1)$
Consider the expansion of${(1 - x)^{ - \dfrac{p}{q}}}$
\[{(1 - x)^{ - \dfrac{p}{q}}} = 1 + \dfrac{p}{{1!}}(\dfrac{x}{q}) + \dfrac{{p(p + q)}}{{2!}}{(\dfrac{x}{q})^2} + ....\]
Comparing (1) with the expansion, we get $p = 3$ and $p + q = 5$.
$ \Rightarrow q = 2$.
Also $\dfrac{x}{q} = \dfrac{1}{4} \Rightarrow x = \dfrac{1}{2}$
Substituting these values in the RHS of equation (1), we get
\[1 + S = {(1 - \dfrac{1}{2})^{ - \dfrac{3}{2}}} = {(\dfrac{1}{2})^{ - \dfrac{3}{2}}} = {(2)^{\dfrac{3}{2}}} = 2\sqrt 2 \]
Now, subtract 1 from both the sides to get S.
Therefore, $S = 2\sqrt 2 - 1$
That is, the sum of the given series is $S = 2\sqrt 2 - 1$.


Note: For any real number x such that$\left| x \right| < 1$ and rational number n, the binomial expansion of ${(1 + x)^n}$ is given by
${(1 + x)^n} = 1 + nx + \dfrac{{n(n - 1)}}{{2!}}{x^2} + .... + \dfrac{{n(n - 1)(n - 2)....(n - r + 1)}}{{r!}}{x^r} + .....$
Where $\dfrac{{n(n - 1)(n - 2)....(n - r + 1)}}{{r!}}$is the coefficient of the\[{r^{th}}\]term of the series