Question

Sendo n um número natural, a expressão $\frac{ (2^{n+1} + 2^{n+2} ) . ( 3^{n+2} - 3^{n-1}) }{ 6^{n+2} }$ é igual a:
a) 1
b) $3^{n}$
c) $2^{n}$
d) $6^{n}$
e) 6

Answer 1

$\dfrac{\left(2^{n+1}+2^{n+2}\right)\cdot\left(3^{n+2}-3^{n-1}\right)}{6^{n+2}}=\dfrac{\left2^{n+1}(1+2\right)\cdot3^{n-1}\left(3^{3}-1\right)}{2^{n+2}\cdot3^{n+2}}=\\\\\\\dfrac{\left2^{n+1}\cdot3^{n}\cdot26}{2^{n+2}\cdot3^{n+2}}=\dfrac{\left2^{n+1}\cdot3^{n}\cdot2\cdot13}{2^{n+2}\cdot3^{n+2}}=\dfrac{\left2^{n+2}\cdot3^{n}\cdot13}{2^{n+2}\cdot3^{n+2}}=\dfrac{3^{n}\cdot13}{3^{n+2}}=\\\\\\\dfrac{13}{3^2}=\bold{\dfrac{13}{9}}$