Question

Aflati x astfel incat $x*3^{2014}= (3^{2014}-1):(1+ \frac{1}{3} + \frac{1}{ 3^{2} } +...+ \frac{1}{ 3^{2013} } ).$ Multumesc.

Answer 1

Consideram suma $S=1+3+ 3^{2} +...+ 3^{n}$.
$3S=3+ 3^{2} +3^{3} +...+ 3^{n+1}$
$3S-S=(3+ 3^{2}+ 3^{3}+...+ 3^{n+1})-(1+3+ 3^{2} +...+ 3^{n}) <=> \\ <=>2S= 3^{n+1}-1 => S= \frac{ 3^{n+1}-1 }{2} .$

$1+ \frac{1}{3} + \frac{1}{3^{2} }+...+ \frac{1}{ 3^{2013} } = \frac{3^{2013}+ 3^{2012}+ 3^{2011}+...+1 }{3^{2013}}= \frac{ \frac{ 3^{2014}-1 }{2} }{ 3^{2013} } = \frac{ 3^{2014}-1}{2* 3^{2013} }$

Inlocuind in ecuatia initiala, obtinem: 
$x* 3^{2014}=( 3^{2014} -1): \frac{ 3^{2014}-1 }{2* 3^{2013} } <=> \\ <=>x* 3^{2014}=( 3^{2014}-1)* \frac{2* 3^{2013} }{ 3^{2014}-1 } <=> \\ <=>x* 3^{2014}=2* 3^{2013} =>3x=2=>x= \frac{2}{3}$