$P_{(x)} = [3+(2x^2-x^{-1})]^{2012}$
$=\sum 3^{2012-k}.C_{2012}^k.(2x^2-x^{-1})^k$
$=\sum 3^{2012-k}.C_{2012}^k.\sum 2^h.C_k^h.x^{2k-2h}.x^{-h}.(-1)^h$
$=\sum.\sum 3^{2012-k}$$C_{2012}^k.2^h.C_k^h.x^{2k-3h}.(-1))^h$
$2k-3h=1001 \Rightarrow k=3t+499 ; h=2t-1$
$0 \leq k \leq 2012 \Rightarrow -166 \leq t \leq 504$
$0 \leq h \leq k \Leftrightarrow t \geq 1$
$\Rightarrow 1 \leq t \leq 504$
Hệ số $x^{1001}$ trong khai triển trên là :
$\sum $$3^{1513-3t}.C_{2012}^{3t+499}$$.2^{2t-1}.C_{3t+499}^{2t-1}.(-1)$
với $t = \overline{1;504}$