Question

Solve:

$\frac{7}{5}(1+\frac{1}{{10}^2}+\frac{1.3}{1.2}.\frac{1}{{10}^4}+\frac{\mathrm{1.3.5}}{\mathrm{1.2.3}}.\frac{1}{{10}^6}+\dots \dots \infty )=$

• A. $\sqrt{2}$
• B. $2\sqrt{2}$
• C. $2^{1/3}$
• D. $\sqrt{\frac{2}{3}}$

Answer 1

Answer: (A)

$\sqrt{2}$

Given,

$\frac{7}{5}(1+\frac{1}{{10}^2}+\frac{1.3}{1.2}.\frac{1}{{10}^4}+\frac{\mathrm{1.3.5}}{\mathrm{1.2.3}}.\frac{1}{{10}^6}+\dots \dots \infty )$

We have,

$(1-x{)}^{-n}=1-n(-x)+\frac{-n(-n-1)x^2}{2!}...\infty$ $=1+nx+\frac{n(n+1)x^2}{2!}...\infty$

Comparing the coefficients with that given in the question we get.
$\Rightarrow$ $nx=\frac{1}{100}$ ...(i)

$\Rightarrow$ $\frac{n(n+1)x^2}{2}=\frac{3}{{2.10}^4}$ ...(ii)

Substituting the value of nx from Eq i in Eq ii we get
$\Rightarrow$ $n{x}^2(n+1)=\frac{3}{{10}^4}$

$\Rightarrow$ $nx(n+1)x=\frac{3}{{10}^4}$

$\Rightarrow$ $(n+1)x\left(\frac{1}{100}\right)=\frac{3}{{10}^4}$

$\Rightarrow$ $nx=\frac{3}{100}-x$

$\Rightarrow$ $\frac{1}{100}=\frac{3}{100}-x$

$\Rightarrow$ $x=\frac{2}{100}$

Since,

$nx=\frac{1}{100}$ $\Rightarrow$ $n=\frac{1}{2}$

Substituting in the original equation the values of n and x we get
$\Rightarrow$ $\frac{7}{5}{(1-\frac{2}{100})}^{\frac{-1}{2}}$

$=\frac{7}{5}{(1-\frac{1}{50})}^{\frac{-1}{2}}$

$=\frac{7}{5}{\left(\frac{49}{50}\right)}^{\frac{-1}{2}}$

$=\frac{7}{5}{\left(\frac{50}{49}\right)}^{\frac{1}{2}}$

$=\frac{7}{5}\left(\frac{5}{7}\right)\sqrt{2}$

$=\sqrt{2}$