+0

# stumper

+1
220
3
+597

Hi good people,

$${{54^n}-18^{n-1}.3^{n+1}} \over{(3^{n-1})}^2.6^n$$

Thank you all very very much, I honestly do appreciate your time...

Mar 5, 2019

#1
+23835
+4

$$\mathbf{\large{ \dfrac{ 54^n-18^{n-1}\cdot 3^{n+1} } { {(3^{n-1})}^2\cdot 6^n } }}$$

$$\begin{array}{|rcll|} \hline &&\mathbf{ \dfrac{ 54^n-18^{n-1}\cdot 3^{n+1} } { {(3^{n-1})}^2\cdot 6^n } } \\\\ &=& \dfrac{ 54^n-18^n\cdot 18^{-1}\cdot 3^n\cdot 3^1 } { { 3^{(n-1)\cdot 2} } \cdot 6^n } \\\\ &=& \dfrac{ 54^n-18^n\cdot 3^n\cdot \frac{3}{18}} { {3^{2n-2}}\cdot 6^n } \\\\ &=& \dfrac{ 54^n-(18\cdot 3)^n\cdot \frac{1}{6} } { {3^{2n}\cdot 3^{-2} }\cdot 6^n } \\\\ &=& \dfrac{ 54^n-54^n\cdot \frac{1}{6} } { \left(3^2\right)^n\cdot \frac{1}{9} \cdot 6^n } \\\\ &=& \dfrac{ 54^n \cdot \left(1- \frac{1}{6}\right) } { 9^n\cdot 6^n \cdot \frac{1}{9} } \\\\ &=& \dfrac{ 54^n \cdot \frac{5}{6} } { \left(6\cdot 9\right)^n \cdot \frac{1}{9} } \\\\ &=& \dfrac{ 54^n \cdot \frac{5}{6} } { 54^n \cdot \frac{1}{9} } \\\\ &=& \dfrac{ \frac{5}{6} } { \frac{1}{9} } \\\\ &=& \dfrac{5}{6} \cdot \dfrac{9}{1} \\\\ &=& \dfrac{5\cdot 3}{2} \\\\ &\mathbf{=}& \mathbf{\dfrac{15}{2}} \\ \hline \end{array}$$

Mar 5, 2019
#2
+597
+2

Heureka,

life saver, I did not come close to this in 3 attempts!!..have a very very blessed day!!!..thank you...

juriemagic  Mar 5, 2019
#3
+23835
+3

Thank you, JM

heureka  Mar 5, 2019