If x=2.53.6−2.5.83.6.9(25)+2.5.8.113.6.9.12(25)2−......∞ , then 72(12x+55)3=

x=2⋅52!(13)2−2⋅5⋅833⋅2!(25)+2⋅5⋅8⋅1134⋅4!(25)2+.......∞ 425x=2⋅532⋅2!(25)2−2⋅5⋅8⋅1133⋅3!(25)3+.............∞ ⇒4x25+1115=(1+25)−23 ⇒12x+5575=(57)23 ⇒72(12x+55)3=753⋅52=33⋅58

If x=2.53.6−2.5.83.6.9(25)+2.5.8.113.6.9.12(25)2−......∞ , then 72(12x+55)3=

x=2⋅52!(13)2−2⋅5⋅833⋅2!(25)+2⋅5⋅8⋅1134⋅4!(25)2+.......∞ 425x=2⋅532⋅2!(25)2−2⋅5⋅8⋅1133⋅3!(25)3+.............∞ ⇒4x25+1115=(1+25)−23 ⇒12x+5575=(57)23 ⇒72(12x+55)3=753⋅52=33⋅58

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If $x = \frac{2.5}{3.6} - \frac{2.5.8}{3.6.9} (\frac{2}{5}) + \frac{2.5.8.11}{3.6.9.12} {(\frac{2}{5})}^{2} - ...... \infty$ , then $7^{2} {(12 x + 55)}^{3} =$

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$3^{8} 5^{3}$

$3^{8} 5^{5}$

$3^{3} 5^{5}$

$3^{3} 5^{8}$

answer is D.

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Detailed Solution

$x = \frac{2 \cdot 5}{2!} {(\frac{1}{3})}^{2} - \frac{2 \cdot 5 \cdot 8}{3^{3} \cdot 2!} (\frac{2}{5}) + \frac{2 \cdot 5 \cdot 8 \cdot 11}{3^{4} \cdot 4!} {(\frac{2}{5})}^{2} + ....... \infty \frac{4}{25} x = \frac{2 \cdot 5}{3^{2} \cdot 2!} {(\frac{2}{5})}^{2} - \frac{2 \cdot 5 \cdot 8 \cdot 11}{3^{3} \cdot 3!} {(\frac{2}{5})}^{3} + ............. \infty \Rightarrow \frac{4 x}{25} + \frac{11}{15} = {(1 + \frac{2}{5})}^{- \frac{2}{3}} \Rightarrow \frac{12 x + 55}{75} = {(\frac{5}{7})}^{\frac{2}{3}} \Rightarrow 7^{2} {(12 x + 55)}^{3} = 75^{3} \cdot 5^{2} = 3^{3} \cdot 5^{8}$