Thank you for reporting, we will resolve it shortly
Q.
$ 1+\frac{2}{4}+\frac{2.5}{4.8}+\frac{2.5.8}{4.8.12}+\frac{2.5.8.11}{4.8.12.16}+... $ is equal to
Bihar CECEBihar CECE 2008
Solution:
Let $S=1+\frac{2}{4}+\frac{2}{4} \cdot \frac{5}{8}+\frac{2.5 .8}{4.8 .12}+\ldots$
On comparing with
$(1+x)^{n}=1+n x+\frac{n(n-1)}{2 !} x^{2}+\ldots .$,
we get $n x=\frac{2}{4}\,\,\,...(i)$
and $\frac{n(n-1)}{2 !} x^{2}=\frac{2.5}{4.8}\,\,\,...(ii)$
From Eqs. (i) and (ii)
$\frac{\frac{n(n-1)}{2 !} x^{2}}{n^{2} x^{2}}=\frac{\frac{2.5}{4.8}}{\frac{2.2}{4.2}}$
$\Rightarrow \frac{n-1}{n}=\frac{5}{2}$
$\Rightarrow 2 n-2=5 n$
$\Rightarrow n=-\frac{2}{3}$
On putting the value of $n$ in Eq. (i), we get
$-\frac{2}{3} x=\frac{2}{4}$
$\Rightarrow x=-\frac{3}{4}$
$\therefore S=\left(1+x^{2}\right)^{n}=\left(1-\frac{3}{4}\right)^{-2 / 3}=\left(\frac{1}{4}\right)^{-2 / 3}=\sqrt[3]{16}$