Let the given series be identical with
$(1+x)^{n}=1+n x+\frac{n(n-1)}{1 \times 2} x^{2}+\cdots \infty$
or $ n x=\frac{1}{4}$ or $n^{2} x^{2}=\frac{1}{16}$
Also, $\frac{n(n-1)}{2} x^{2}=\frac{3}{32}$ or $\frac{2 n}{n-1}=\frac{\frac{1}{16}}{\frac{3}{32}}=\frac{2}{3}$
or $3 n=n-1$
or$ 2 n=-1 $
or $ n=-\frac{1}{2} $
$\Rightarrow x=-\frac{1}{2}$
$ \therefore $ Required sum$=\left(1-\frac{1}{2}\right)^{-\frac{1}{2}}=\left(\frac{1}{2}\right)^{-\frac{1}{2}} $
$=(2)^{\frac{1}{2}}=\sqrt{2} $