Evaluate $\sum_{k=400}^{2000} \frac {2^{3-4k}} {8^{2k+3}}$ 
Evaluate $$\sum_{k=400}^{2000} \frac {2^{3-4k}} {8^{2k+3}}$$

So  far, I was able to get to
$$\frac{1}{64}\sum_{k=400}^{2000} \frac {1} {8^{2k}\cdot2^{4k}}$$
And then I'm completely stuck. 
 A: Hint: $8^{2k}\cdot 2^{4k} = (64\cdot 16)^k = 1,024^k$
A: Note $8=2^3$, so 
$$\sum_{k=400}^{2000} \frac {2^{3-4k}} {8^{2k+3}}=\sum_{k=400}^{2000} \frac {2^{3-4k}} {2^{3(2k+3)}}=\sum_{k=400}^{2000} \frac {2^{3-4k}} {2^{6k+9}}=\sum_{k=400}^{2000}2^{-10k-6}$$
and this is a geometric sequence with first term $2^{-4006}$ and ratio $2^{-10}$. Solve it using the formula
$$S=a\left(\frac{r^n-1}{r-1}\right)$$
where $a$ is the first term, $n$ is the number of terms, and $r$ is the common ratio. 
A: $$\begin{align}
\large \sum_{k=400}^{2000} \frac {2^{3-4k}} {8^{2k+3}}
&=\large\sum_{k=400}^{2000}2^{3-4k-3(2k+3)}\\
&=\large\sum_{k=400}^{2000}2^{-10k-6}\\
&=\large\frac 1{2^6}\sum_{k=400}^{2000}\frac 1{2^{10k}}\\
&=\large\frac 1{2^6}\sum_{k=400}^{2000}{u^{k}}\quad \text{where $u=\large \frac 1{2^{10}}$}\\
&=\large\frac 1{2^6}\frac {u^{400}(1-u^{1601})}{1-u}\\
&=\large\large\frac 1{2^6}\frac {\frac 1{2^{4000}}{(1-\frac 1{2^{16010}})}}{1-\frac 1{2^{10}}}\\
&=\large\frac {\frac 1{2^{3996}}{(1-\frac 1{2^{16010}})}}{2^{10}-1}\\
\end{align}$$
...it does look rather messy...
