Question

Pls answer. The correct answer gets marked as brainliest.

Answer 1

PROVED !!

Answer:

$\LARGE\bold{LHS}$

$\LARGE{ = \frac{ {3}^{n} + {3}^{n - 1} }{ {3}^{n + 1} - {3}^{n} } }$

$\LARGE{ = \frac{ {3}^{(n)} + {3}^{(n - 1)} }{ {3}^{(n + 1)} - {3}^{(n)} } }$

$\LARGE{ = \frac{ {3}^{n} + \frac{ {3}^{n} }{ {3}^{1} } }{( {3}^{n} \times {3}^{1} ) - {3}^{n} } }$—————$\Large{ \left[ \because \frac{ {x}^{a} }{ {x}^{b} } = {x}^{(a - b)} \: \: and \: \: {x}^{a} \times {x}^{b} = {x}^{(a + b)} \right] }$

$\LARGE{ = \frac{ {3}^{n} + \frac{ {3}^{n} }{3} }{( {3}^{n} \times 3) - {3}^{n} } }$

$\LARGE{ = \frac{ {3}^{n} \left( 1 + \frac{1}{3} \right) }{ {3}^{n} (3 - 1)} }$——————[ taking ${3}^{n}$ as common both in numerator and denominator ]

$\LARGE{ = \frac{ \cancel{ {3}^{n} } \left( 1 + \frac{1}{3} \right) }{ \cancel{ {3}^{n} } (3 - 1)} }$

$\LARGE{ = \frac{ \left( 1 + \frac{1}{3} \right) }{(3 - 1)} }$

$\LARGE{ = \frac{ \left( \frac{3 + 1}{3} \right) }{2} }$

$\LARGE{ = \frac{ \left( \frac{4}{3} \right) }{2} }$

$\LARGE{ = \frac{4}{3} \times \frac{1}{2} }$

$\LARGE{ = \frac{ \cancel{4} \: {}^{2} }{3} \times \frac{1}{ \cancel{2} \: _1} }$

$\LARGE{ = \frac{2}{3} \times \frac{1}{1} }$

$\LARGE{ = \frac{2}{3} }$

$\LARGE\bold{ = RHS}$