Simplifying the generalization of a summation I guess I am having trouble with the algebra, could someone walk me through this probably simple simplification?
I currently have $\frac{x^2\sqrt3}{4} +  $$\sum_{i=0}^n 3\cdot4^i \frac{\frac{x}{3^{i-1}}^2 \sqrt3}{4} $
I am pretty sure that I am supposed to be getting:
$$\rm {area}={\sqrt{3}\over 4}x^2+\sum_{i=0}^{\infty}{1\over4\sqrt{3}}x^2({4\over9})^i$$
Can someone help me understand how to get there from what I have? 
 A: \begin{align}
\sum_{i=0}^n 3\cdot4^i \frac{\left(\frac{x}{3^{i-1}}\right)^2 \sqrt3}{4}&=\frac{3x^2\sqrt{3}}{4}\sum_{i=0}^n3^{-2(i-1)}\cdot4^i\tag{1}\\
&=\frac{3^2\cdot3x^2\sqrt{3}}{4}\sum_{i=0}^n(3^{-2})^i\cdot4^{i}\tag{2}\\
&=\frac{3^3x^2\sqrt{3}}{4}\sum_{i=0}^n(3^{-2}\cdot4)^i\tag{3}\\
&=\frac{3^3x^2\sqrt{3}}{4}\sum_{i=0}^n\left(\frac{4}{9}\right)^i\tag{4}
\end{align}
Explanations:
$(1)$ Take out the terms not depending on the summation index $i$ and use the following general properties of exponentiation:


*

*$(\frac{a}{b})^c=\frac{a^c}{b^c}$

*$\frac{1}{a^b}=a^{-b}$

*$(a^b)^c=a^{bc}$


$(2)$ Note that $3^{-2(i-1)}=3^{-2i+2}$ and use the following general property of exponentation:


*

*$a^{b+c}=a^b\cdot a^c$


to take the constant multiplicative term (constant with respect to the index of summation $i$) $3^2$ out of the sum. Also, use the third property listed to get $3^{-2i}=(3^{-2})^i$.
$(3)$ Use the following general property of exponentiation:


*

*$(ab)^c=a^c\cdot b^c$


$(4)$ Now we have succeeded in simplifying the sum in order to recognize a geometric sum. Hopefully you can take it from here.
