Question

`\frac { 18 ^ { n + 1 } * 3 ^ { 1 - n } } { 2 ^ { n - 1 } }`

Somebody please help to simplify it.​

Answer 1

Explanation:

`\sf \large \boldsymbol{} 1) \ a^n \cdot b^n= (ab) ^n \\\\2) \ a^n\cdot a^m= a^(m* n ) \\\\\\\displaystyle (18^(n+1)\cdot 3^(1-n ))/(2^(n-1)) =((2\cdot 9)^(n+1)\cdot 3^(1-n ))/(2^(n-1)) =(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) }`

Answer 2

`4 * {3}^(n + 3)`

Explanation:

`\frac{ {18}^(n + 1) * {3}^(1 - n)}{ {2}^(n - 1) }`

`\frac{ {3}^(2n + 2) * {2}^(n + 1) * {3}^(1 - n) }{ {2}^(n - 1) }`

`{3}^(2n + 2) * {2}^(2) * {3}^(1 - n)`

`{3}^(2n + 2) * 4 * {3}^(1 - n)`

`= 4 * {3}^(n + 3)`