Discussion Forum

1)

If   $x=\frac{2.5}{3.6}-\frac{2.5.8}{3.6.9}\left(\frac{2}{5}\right)+\frac{2.5.8.11}{3.6.9.12}\left(\frac{2}{5}\right)^{2}-... \infty$  then

$7^{2}(12x+55)^{3}$=

Answer:

Option D

Explanation:

$x=\frac{2.5}{3.6}-\frac{2.5.8}{3.6.9}\left(\frac{2}{5}\right)+\frac{2.5.8.11}{3.6.9.12}\left(\frac{2}{5}\right)^{2}-... \infty$

$=\frac{2.5}{3^{2}.2!}-\frac{2.5.8}{3^{3}.3!}\left(\frac{2}{5}\right)+\frac{2.5.8.11}{3^{4}.4!}\left(\frac{2}{5}\right)^{2}$

mulitply with $(\frac{2}{5})^{2}$

$\frac{4}{25}x=\frac{2.5}{3^{2}.2!}\left(\frac{2}{5}\right)^{2}-\frac{2.5.8}{3^{3}.3!}\left(\frac{2}{5}\right)^{3}+......\infty$

$\frac{4}{25}x-\frac{2.5}{3^{}.1!}\left(\frac{2}{5}\right)^{}+1=1-\frac{2}{3^{}.1!}\left(\frac{2}{5}\right)^{}+\frac{2.5}{3^{2}.2!}\left(\frac{2}{5}\right)^{2}-\frac{2.5.8}{3^{3}.3!}\left(\frac{2}{5}\right)^{3}+......$

$\frac{4x}{25}-\frac{4}{15}+1=\left(1+\frac{2}{5}\right)^{-2/3}$

$\Rightarrow$    $\frac{4x}{25}+\frac{11}{15}=\frac{1}{5}\left(\frac{7}{5}\right)^{2/3}\Rightarrow\left(\frac{4x}{5}+\frac{11}{3}\right)=\left(\frac{7}{5}\right)^{-2/3}$

$\Rightarrow$    $\frac{1}{5}\left(\frac{12x+55}{15}\right)=\left(\frac{7}{5}\right)^{-2/3}\Rightarrow\frac{12x+55}{75}=\left(\frac{5}{7}\right)^{2/3}$

cube both sides,

$\Rightarrow \frac{(12x+55)^{3}}{(79)^{3}}=\frac{5^{2}}{7^{2}}\Rightarrow 7^{2}(12x+55)^{3}=75^{3}.5^{2}$

$=3^{3}.5^{6}.5^{2}=3^{3}.5^{8}$