(15) Let $\left\{a_{n}\right\}$ be a geometric sequence, and each term is greater than 1, then $\lg a_{1} \lg a_{2012} \sum_{i=1}^{20111} \frac{1}{\lg a_{i} \lg a_{i+1}}$ has the value $\qquad$.
152011 Prompt: When the common ratio is 1, then
$$
\lg a_{1} \lg a_{2012} \sum_{i=1}^{2011} \frac{1}{\lg a_{i} \lg a_{i+1}}=2011 \text {. }
$$

When the common ratio is $q \neq 1$, then
$$
\begin{aligned}
& \lg a_{1} \lg a_{2012} \sum_{i=1}^{2011} \frac{1}{\lg a_{i} \lg a_{i+1}} \\
= & \frac{\lg a_{1} \lg a_{2012}}{\lg q} \sum_{i=1}^{2011}\left(\frac{1}{\lg a_{i}}-\frac{1}{\lg a_{i+1}}\right) \\
= & \frac{\lg a_{1} \lg a_{2012}}{\lg q}\left(\frac{1}{\lg a_{1}}-\frac{1}{\lg a_{2012}}\right) \\
= & 2011 .
\end{aligned}
$$