Question

$1-\frac{3}{16}+\frac{1.4}{1.2}{\left(\frac{3}{16}\right)}^2-\frac{\mathrm{1.4.7}}{\mathrm{1.2.3}}{\left(\frac{3}{16}\right)}^3+\dots \dots =$

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