Tìm số hạng không chứa x trong khai triẻn  : $(\frac{2}{\sqrt[3]{x}}-x^2)^n$
Biết: $\mathbb{C}_{2n+1}^{1}+\mathbb{C}_{2n+1}^{2}+....+\mathbb{C}_{2n+1}^{n}=2^{28}-1$ 

$pt<=>C^0_{2n+1}+C^1_{2n+1}+...+C^n_{2n+1}=2^{28}$

$<>C^0_{2n+1}+C^1_{2n+1}+...+C^n_{2n+1}+C^{n+1}_{2n+1}+...+C^{2n+1}_{2n+1}=2^{29}$(nhân 2 vào hai vế, hai số hạng có tổng số mũ $=2n+1$ thì bằng nhau)


$<=>2^{2n+1}=2^{29}$


$<=>n=14$


$(\frac{2}{\sqrt[3]x}-x^2)^{14}=\sum_{k=0}^{14} C^{k}_{14}(\frac{2}{\sqrt[3]x})^k.(-x^2)^{14-k}$


$=\sum_{k=0}^{14} C^{k}_{14}2^kx^{\frac{-k}{3}}(-1)^{14-k}x^{28-2k}$


$=\sum_{k=0}^{14} C^k_{14}2^k.(-1)^{14-k}x^{28-\frac{7k}{3}}$


để số hạng  thứ $k+1$ không chứa $x$ thì $28-\frac{7k}{3}=0<=>k=12$


Vậy số hạng đó là $C^{12}_{14}.2^{12}$

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