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ikadub [295]
3 years ago
5

D%20%7D%20%7B%202%20%5E%20%7B%20n%20-%201%20%7D%20%7D" id="TexFormula1" title="\frac { 18 ^ { n + 1 } \times 3 ^ { 1 - n } } { 2 ^ { n - 1 } }" alt="\frac { 18 ^ { n + 1 } \times 3 ^ { 1 - n } } { 2 ^ { n - 1 } }" align="absmiddle" class="latex-formula">
Somebody please help to simplify it.​
Mathematics
2 answers:
Rufina [12.5K]3 years ago
8 0

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}

Nostrana [21]3 years ago
6 0

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} }

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