Question

The sum of series $\dfrac{3}{4\times 8}-\dfrac{3\times 5}{4\times 8\times 12}+\dfrac{3\times 5\times 7}{4\times 8\times 12\times 16}-........$ is
A) $\sqrt{\dfrac{3}{2}}-\dfrac{3}{4}$
B) $\sqrt{\dfrac{2}{3}}-\dfrac{3}{4}$
C) $\sqrt{\dfrac{3}{2}}-\dfrac{1}{4}$
D) $\sqrt{\dfrac{2}{3}}-\dfrac{1}{4}$

Answer 1

According to this question infinite series is given that is $S=\dfrac{3}{4\times 8}-\dfrac{3\times 5}{4\times 8\times 12}+\dfrac{3\times 5\times 7}{4\times 8\times 12\times 16}-........$ to solve such types of problems we have to add $1-\dfrac{1}{4}$ on both sides and then apply the formula for series that is ${{(1+x)}^{-1}}=1-x+{{x}^{2}}-{{x}^{3}}.......$ and solve further such types of problems.

Complete step by step answer:
In the question infinite series is given that is $S=\dfrac{3}{4\times 8}-\dfrac{3\times 5}{4\times 8\times 12}+\dfrac{3\times 5\times 7}{4\times 8\times 12\times 16}-........$
To simplify this type of series we have to add $1-\dfrac{1}{4}$ on both sides
By adding on both side of this equation we get:
$1-\dfrac{1}{4}+S=1-\dfrac{1}{4}+\dfrac{3}{4\times 8}-\dfrac{3\times 5}{4\times 8\times 12}+\dfrac{3\times 5\times 7}{4\times 8\times 12\times 16}-........$
This infinite series we have to represent in a standard formula for that we have to simplify it further we get:
$\dfrac{3}{4}+S=1-\dfrac{1}{4}+\dfrac{3}{4\times 8}-\dfrac{3\times 5}{4\times 8\times 12}+\dfrac{3\times 5\times 7}{4\times 8\times 12\times 16}-........$
Above infinite series can also be written as
$\dfrac{3}{4}+S=1-1\times \left( \dfrac{1}{4} \right)+\dfrac{1\times 3}{1\times 2}{{\left( \dfrac{1}{4} \right)}^{2}}-\dfrac{1\times 3\times 5}{1\times 2\times 3}{{\left( \dfrac{1}{4} \right)}^{3}}+\dfrac{1\times 3\times 5\times 7}{1\times 2\times 3}{{\left( \dfrac{1}{4} \right)}^{4}}-........$
To simplify the further equation by considering value such as p=1$$$$q=2$$$$x=\dfrac{1}{2}
Before substituting the value we have to more simplify and separate the term so that substitutions become easier.
$\dfrac{3}{4}+S=1-\dfrac{1}{1}\times \left( \dfrac{1}{2\times 2} \right)+\dfrac{1\times (1+2)}{1\times 2}{{\left( \dfrac{1}{2\times 2} \right)}^{2}}-\dfrac{1\times (1+2)\times (1+2(2))}{1\times 2\times 3}{{\left( \dfrac{1}{2\times 2} \right)}^{3}}+........$
Now, you can substitute the value in the above series
$\dfrac{3}{4}+S=1-\dfrac{p}{1}\times \left( \dfrac{x}{q} \right)+\dfrac{p\times (p+q)}{1\times 2}{{\left( \dfrac{x}{q} \right)}^{2}}-\dfrac{p\times (p+q)\times (p+2q)}{1\times 2\times 3}{{\left( \dfrac{x}{q} \right)}^{3}}+........$
If you carefully observe the series then this type of series will be for ${{(1+x)}^{-m}}$
${{(1+x)}^{-m}}{{=}^{m}}{{C}_{0}}{{-}^{m}}{{C}_{1}}x{{+}^{m}}{{C}_{1}}{{x}^{2}}+........$ (From binomial distribution)
Apply this type of series in the above equation we get:
$\dfrac{3}{4}+S={{(1+x)}^{-\dfrac{p}{q}}}$
After rearranging the term you will get the value of S
$S={{(1+x)}^{-\dfrac{p}{q}}}-\dfrac{3}{4}$
Again substitute the value of $p=1$,$q=2$ and $x=\dfrac{1}{2}$ so we get the answer accurately
$S={{\left( 1+\dfrac{1}{2} \right)}^{-\dfrac{1}{2}}}-\dfrac{3}{4}$
After simplifying this we get:
$S={{\left( \dfrac{3}{2} \right)}^{-\dfrac{1}{2}}}-\dfrac{3}{4}$
This equation can also be written as
$S=\sqrt{\dfrac{2}{3}}-\dfrac{3}{4}$
Therefore, the sum of the infinite series will be $S=\sqrt{\dfrac{2}{3}}-\dfrac{3}{4}$.
So, the correct option is “option (B)”.

Note:
Here, in this particular problem to make the series simpler we have to add $1-\dfrac{1}{4}$ on both sides.
Take little care while substituting the value and to avoid confusion always consider like p=1$$$$q=2$$$$x=\dfrac{1}{2} so, that substitution and it’s easy to detect the formula. So, in this way we can solve and the above solution can be preferred for such types of problems.