The sum of the series 
$$\frac{1}{\log_24}+\frac{1}{\log_44} + \frac{1}{\log_84}.....\frac{1}{\log_{2^n}4}$$

MY SOLUTION
We can write it as 
$$\frac{\log2}{\log4} + \frac{\log4}{\log4} + ....\frac{\log 2^n}{\log4}$$
$$=\frac{1}{\log4}\left[\log 2 + \log 4....\log 2^n\right]$$
$$=\frac{1}{\log 4}\left [\log(2.4.8....2^n)\right]$$
$$\frac{1}{\log 4}[log(2^n.2^{\frac{(n)(n+1)}{2}}
)]$$
$$\frac{2n+n^2+n}{4}$$
But the answer is $\frac{n^2+n}{4}$
 A: You are almost right, we have that
$$\dots=\frac{1}{\log 4}\left [\log(2\cdot 4\cdot 8\cdot \ldots \cdot 2^n)\right]=\frac{1}{\log 4}\left [\log(2\cdot 2^2\cdot 2^3\cdot \ldots \cdot 2^n)\right]=$$
$$=\frac{1}{\log 4}\left [\log(2^{1+2+3+\dots+n}
)\right]=\frac{1}{\log 4}\left [\log(2^{\frac{n(n+1)}2})\right]=\frac{n(n+1)}2\frac{\log 2}{\log 4}=\frac{n(n+1)}4$$
indeed $\frac{\log 2}{\log 4}=\frac12 \frac{2\log 2}{\log 4}=\frac12 \frac{\log 4}{\log 4}=\frac12$.
A: The series can be written by reversing the bases,
$$\log_{4} (2)+\log_{4} (4) +\log_4 (8)+\cdots$$
$$\log_{4}(2.4.8...)=\log_{4}(2^{\frac{n (n+1)}{2}}) $$
Which is 
$$\frac {n^2+n}{4} $$

There is an error in the product $2\cdot4\cdot8\cdots$ you took in your solution .
It can be written as $$2\cdot2^2\cdot2^3...=2^{1+2+3+...} $$
Which is $$2^{\frac{n (n+1)}{2}}$$
A: $\dfrac{1}{2\log 2}[\log 2(1+2+3+....n)]=$
$\dfrac{n(n+1)}{2\cdot 2}$;
Used: 
1)$\log 4= \log 2^2=2\log/2$;
2)$\log 2+\log 2^2+.......\log 2^n=$
$\log 2 +2\log 2 +....+n\log 2=$
$\log 2(1+2+3+.....n)=$
$\log 2\dfrac{n(n+1)}{2}$.
