$$3^{\log_4^n} + \sum_{a=0}^{(\log_4^n)-1}3^a$$
Attempt: using formula $\sum_{k=0}^{n}r^k = \frac{{1-r^{n+1}}}{1-r}$ when $ r > 1$, I get: $$3^{\log_4^n} + \frac{{1-3^{(\log_4^n-1)+1}}}{1-3} = 3^{\log_4^n} + \frac{{1-3^{\log_4^n}}}{-2}$$
Supposedly, this should end up becoming:
$\frac{3^{\log_4^{4n}}-1}{2}$
But I'm stuck. Somehow $\frac{1-3^{\log_4^n}}{-2}$ is multiplied by $-1$ and $3^{\log_4^n}$ is eliminated completely?
Your last term is $3^{\log_4^n}+\frac{1-3^{\log_4^n}}{-2}$. Computing common denominator gives $$\frac{-2\times 3^{\log_4^n}+1-3^{\log_4^n}}{-2}=\frac{-3\times 3^{\log_4^n}+1}{-2}=\frac{- 3^{1+\log_4^n}+1}{-2}=\frac{- 3^{\log^4_4+\log_4^n}+1}{-2}$$ Using properties of $\log$ we have $$\frac{- 3^{\log_4^{4n}}+1}{-2}=\frac{3^{\log_4^{4n}}-1}{2}$$