Question

$\frac { 18 ^ { n + 1 } \times 3 ^ { 1 - n } } { 2 ^ { n - 1 } }$
Somebody please help to simplify it.​

Answer 1

Answer:

$4 \times {3}^{n + 3}$

Step-by-step explanation:

$\frac{ {18}^{n + 1} \times {3}^{1 - n}}{ {2}^{n - 1} }$

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

${3}^{2n + 2} \times {2}^{2} \times {3}^{1 - n}$

${3}^{2n + 2} \times 4 \times {3}^{1 - n}$

$= 4 \times {3}^{n + 3}$

Answer 2

Answer:

Step-by-step explanation:

$\sf \large \boldsymbol{} 1) \ a^n \cdot b^n= (ab) ^n \\\\2) \ a^n\cdot a^m= a^{m\times n } \\\\\\\displaystyle \frac{18^{n+1}\cdot 3^{1-n }}{2^{n-1}} =\frac{(2\cdot 9)^{n+1}\cdot 3^{1-n }}{2^{n-1}} =\frac{2^{n+1}\cdot 9 ^{n+1}\cdot 3^{1-n }}{2^{n-1}} = \\\\\\\frac{\boldsymbol {\sf 2^n\!\!\!\!/}\cdot 2\cdot (3)^{2(n+1)}\cdot 3^{1-n}}{\boldsymbol {\sf 2^n \!\!\!\!/}: 2} =2\cdot 2\cdot 3^{2n+2}\cdot 3^{1-n }=4\cdot 3^{2n+2-n+1}= \\\\\\ \boxed {\sf 4\cdot 3^{n+3} }$