Question

$1-\frac{3}{16}+\frac{1.4}{1.2}\left(\frac{3}{16}\right)^{2}-\frac{1.4 .7}{1.2 .3}\left(\frac{3}{16}\right)^{3}+\ldots \ldots=$

• A. $\left(\frac{15}{6}\right)^{\frac{3}{8}}$
• B. $\left(\frac{4}{5}\right)^{\frac{2}{3}}$
• C. $\left(\frac{7}{4}\right)^{\frac{1}{16}}$
• D. $\left(\frac{4}{15}\right)^{-\frac{2}{5}}$

Answer 1

Answer: (B)

We have,
$1-\frac{3}{16}+\frac{1 \cdot 4}{1 \cdot 2}\left(\frac{3}{16}\right)^{2}-\frac{1 \cdot 4 \cdot 7}{1 \cdot 2 \cdot 3}\left(\frac{3}{16}\right)^{3}+\ldots=$
We know that,
$(1-x)^{n}=1-n x+\frac{n(n-1)}{2 !} x^{2}-\frac{n(n-1)(n-2)}{3 !} x^{3} $
Let $(1-x)^{n}=1-\frac{3}{16}+\frac{1 \cdot 4}{1 \cdot 2}\left(\frac{3}{16}\right)^{2}-\frac{1 \cdot 4 \cdot 7}{1 \cdot 2 \cdot 3}\left(\frac{3}{16}\right)^{3} \cdot s$
$\therefore n x=\frac{3}{16}$ and $ \frac{n(n-1)}{2 !} x^{2}=\frac{1 \cdot 4}{2 !}\left(\frac{3}{16}\right)^{2} $
$n x=\frac{3}{16} $ and $ n(n-1) x^{2}=4\left(\frac{3}{16}\right)^{2} $
$n^{2} \cdot x^{2}=\frac{3^{2}}{16} $ and $n(n-1) x^{2}=4\left(\frac{3}{16}\right)^{2} $
$n(n-1) x^{2}=4 n^{2} x^{2} $
$n^{2}-n=4 n^{2} $
$n-1=4 n $
$\Rightarrow n=-\frac{1}{3} $
$x=\frac{3}{16} \times-3=\frac{-9}{16} $
$(1-x)^{n}=\left(1+\frac{9}{16}\right)^{-1 / 3}$
$=\left(\frac{25}{16}\right)^{-1 / 3}$
$=\left(\frac{4}{5}\right)^{2 / 3}$