[Toán 11] Nhị thức Niw ton

N

noinhobinhyen

$P_{(x)} = [3+(2x^2-x^{-1})]^{2012}$

$=\sum 3^{2012-k}.C_{2012}^k.(2x^2-x^{-1})^k$

$=\sum 3^{2012-k}.C_{2012}^k.\sum 2^h.C_k^h.x^{2k-2h}.x^{-h}.(-1)^h$

$=\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 $t = \overline{1;504}$
 
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