Question

Prove that:-
$\frac{ {2}^{n} + {2}^{n - 1} }{ {2}^{n + 1} - {2}^{n} } = \frac{3}{2}$
Plz Fast
And no SPAMS​

Answer 1

Step-by-step explanation:

$2 {}^{n} + 2 {}^{n - 1} \div 2 {}^{n + 1} - 2 {}^{n}$

=

$2 {}^{n} (1 + \frac{1}{2} ) \div 2 {}^{n} (2 - 1)$

=

$\frac{3}{2} \div 1$

=

$\frac{3}{2}$

Answer 2

there u goo

2

n

+2

n−1

÷2

n+1

−2

n

=

2 {}^{n} (1 + \frac{1}{2} ) \div 2 {}^{n} (2 - 1)2

n

(1+

2

1

)÷2

n

(2−1)

=

\frac{3}{2} \div 1

2

3

÷1

=

\frac{3}{2}

2

3