Lots of logarithms

$\begin{aligned} a_{k+1} & = 4a_k+3 \\ a_{k+1} + 1 & = 4a_k+4 \\ \Rightarrow a_{k+1} + 1 & = 4(a_k+1) \\ \Rightarrow a_1 + 1 & = 4 & \Rightarrow \log_2{(a_1 + 1)} = 2 \\ a_2 + 1 & = 4^2 & \Rightarrow \log_2{(a_2 + 1)} = 4 \\ a_3 + 1 & = 4^3 & \Rightarrow \log_2{(a_3 + 1)} = 6 \\ ... & & ... \\ \Rightarrow a_k + 1 & = 4^k & \Rightarrow \log_2{(a_k + 1)} = 2k \end{aligned}$

$\begin{aligned} \sum_{k=1}^{1000} {\frac{1}{\log_2{(a_k + 1)}\log_2{(a_{k+1} + 1)}}} & = \sum_{k=1}^{1000} {\frac{1}{4k(k+1)}} \\ & = \frac{1}{4} \sum_{k=1}^{1000} {\left(\frac{1}{k} - \frac{1}{k+1}\right)} \\ & = \frac{1}{4} \left( \sum_{k=1}^{1000} {\frac{1}{k}} - \sum_{k=2}^{1001} {\frac{1}{k}} \right) \\ & = \frac{1}{4} \left( \frac{1}{1} - \frac{1}{1001} \right) \\ & = \frac{1}{4} \left( \frac{1000}{1001} \right) \\ & = \frac {250}{1001} \end{aligned}$

$\Rightarrow m + n = \boxed{1251}$