Question

If $x=\frac{2\cdot 5}{3\cdot 6}-\frac{2\cdot 5\cdot 8}{3\cdot 6\cdot 9}\left(\frac{2}{5}\right)+\frac{2\cdot 5\cdot 8\cdot 11}{3\cdot 6\cdot 9\cdot 12}{\left(\frac{2}{5}\right)}^2-\dots \infty$ then $7^2(12x+55{)}^3=$

• A. $3^8{5}^3$
• B. $3^8{5}^5$
• C. $3^3{5}^5$
• D. $3^3{5}^8$

Answer 1

Answer: (D)

We have,
$x=\frac{2\cdot 5}{3\cdot 6}-\frac{2\cdot 5\cdot 8}{3\cdot 6\cdot 9}\left(\frac{2}{5}\right)+\frac{2\cdot 5\cdot 8\cdot 11}{3\cdot 6\cdot 9\cdot 12}{\left(\frac{2}{5}\right)}^2+\dots \infty$
$=\frac{2\cdot 5}{3^2\cdot 2!}-\frac{2\cdot 5\cdot 8}{3^3\cdot 3!}\left(\frac{2}{5}\right)+\frac{2\cdot 5\cdot 8\cdot 11}{3^4\cdot 4!}{\left(\frac{2}{5}\right)}^2$
multiply with ${\left(\frac{2}{5}\right)}^2$
$\frac{4}{25}x=\frac{2\cdot 5}{3^2\cdot 2!}{\left(\frac{2}{5}\right)}^2-\frac{2\cdot 5\cdot 8}{3^3\cdot 3!}{\left(\frac{2}{5}\right)}^3+\dots \infty$
$\frac{4}{25}x-\frac{2}{3\cdot 1!}\left(\frac{2}{5}\right)+1=1-\frac{2}{3\cdot 1!}\left(\frac{2}{5}\right)+\frac{2\cdot 5}{3^2\cdot 2!}$
${\left(\frac{2}{5}\right)}^2-\frac{2\cdot 5\cdot 8}{3^3\cdot 3!}{\left(\frac{2}{5}\right)}^3+\dots$
$\frac{4x}{25}-\frac{4}{15}+1={\left(1+\frac{2}{5}\right)}^{-\frac{2}{3}}$
$\Rightarrow \frac{4x}{25}+\frac{11}{15}={\left(\frac{7}{5}\right)}^{-\frac{2}{3}}$
$\Rightarrow \frac{1}{5}\left(\frac{4x}{5}+\frac{11}{3}\right)={\left(\frac{7}{5}\right)}^{-\frac{2}{3}}$
$\Rightarrow \frac{1}{5}\left(\frac{12x+55}{15}\right)={\left(\frac{7}{5}\right)}^{-\frac{2}{3}}$
$\Rightarrow \frac{12x+55}{75}={\left(\frac{5}{7}\right)}^{\frac{2}{3}}$
cube both sides,
$\Rightarrow \frac{(12x+55{)}^3}{(75{)}^3}=\frac{5^2}{7^2}$
$\Rightarrow 7^2(12x+55{)}^3=75^3\cdot 5^2$
$=3^3\cdot 5^6\cdot 5^2=3^3\cdot 5^8$