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  4. If x=(2 ⋅ 5/3 ⋅ 6)-(2 ⋅ 5 ⋅ 8/3 ⋅ 6 ⋅ 9)((2/5))+(2 ⋅ 5 ⋅ 8 ⋅ 11/3 ⋅ 6 ⋅ 9 ⋅ 12)((2/5))2- ldots ∞ then 72(12 x+55)3=

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

TS EAMCET 2018

Solution:

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