MCQ
$1 + \frac{1}{4} + \frac{{1.3}}{{4.8}} + \frac{{1.3.5}}{{4.8.12}} + ........... = $
  • $\sqrt 2 $
  • B
    $\frac{1}{{\sqrt 2 }}$
  • C
    $\sqrt 3 $
  • D
    $\frac{1}{{\sqrt 3 }}$

Answer

Correct option: A.
$\sqrt 2 $
a
(a) Let the given series be identical with the expansion of ${(1 + x)^n}$ $i.e.$ with $1 + nx + \frac{{n(n - 1)}}{{2\,!}}{x^2} + ....;\,|x|\, < \,1$.

Then, $nx = \frac{1}{4}$ and $\frac{{n(n - 1)}}{2}{x^2} = \frac{1}{4}\,.\,\frac{3}{8} = \frac{3}{{32}}$

Solving these two equations for $n$ and $x$. We get $x = - \frac{1}{2}$ and $n = - \frac{1}{2}$.

$ \therefore$ Sum of the given series
= ${(1 + x)^n} = {\left( {1 - \frac{1}{2}} \right)^{ - 1/2}} = {2^{1/2}} = \sqrt {2.} $

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