Question

Prove that :-
$\frac{ {3}^{n} + {3}^{n - 1} }{ {3}^{n + 1} - {3}^{n} } = \frac{2}{3}$
​

Answer 1

Prove that :-

$\frac{ {3}^{n} + {3}^{n - 1} }{ {3}^{n + 1} - {3}^{n} } = \frac{2}{3}$

Explanation:-

we have LHS:-

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

$= \frac{ {3}^{n} + {3}^{n} \div 3 }{ {3}^{n} \times 3 - {3}^{n} }$

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

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

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

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

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

Answer 2

Answer:

________________________________

________________

_________

____

⭐

Solution:-

_____________

⭐Plz refer to attached ⭐

hope helps✊