Tổng S=2019C0 + 2019C3 + 2019C6 + ... + 2019C2019 bằng:

Câu hỏi:

Tổng $S=C_{2019}^0+C_{2019}^3+C_{2019}^6+...+C_{2019}^{2019}$ bằng:

A. $\frac{2^{2019}-2}{3}$.

B. $\frac{2^{2019}+4}{3}$.

C. $\frac{2^{2019}+2}{3}$.

D. $\frac{2^{2019}-4}{3}$.

Trả lời:

Đáp án cần chọn là: A

Ta tìm các số phức z thỏa mãn $z^3=1$ ta có:

$$\begin{gathered} z^3=1\iff z^3-1=0\leftrightarrow \left(z-1\right)\left(z^2+z+1\right)=0 \\ \iff \begin{array}{rcl}z& =& 1\\ z^2+z+1& =& 0\end{array}\iff \begin{array}{rcl}{z}_1& =& 1\\ z_2& =& -\frac{1}{2}+\frac{\sqrt{3}}{2}i\\ z_3& =& -\frac{1}{2}-\frac{\sqrt{3}}{2}i\end{array} \end{gathered}$$

Xét kai triển

$$\begin{gathered} {\left(1+x\right)}^{2019}=\sum _{k=0}^{2019}{C}_{2019}^k{x}^k \\ =C_{2019}^0+C_{2019}^{1}x+C_{2019}^2{x}^2+...+C_{2019}^{2019}{x}^{2019} \end{gathered}$$(*)

Thay $z_2=-\frac{1}{2}+\frac{\sqrt{3}}{2}i$ vào khai triển (*) ta được

$$\begin{gathered} {\left(1-\frac{1}{2}+\frac{\sqrt{3}}{2}i\right)}^{2019}=C_{2019}^0+C_{2019}^1{z}_2+C_{2019}^2.{z_2}^2+...+C_{2019}^{2019}{z_2}^{2019} \\ \iff {\left(\frac{1}{2}+\frac{\sqrt{3}}{2}i\right)}^{2019}=C_{2019}^0+C_{2019}^1{z}_2+C_{2019}^2{z_2}^2+C_{2019}^3+...+C_{2019}^{2019} \\ \iff -1=\left(C_{2019}^0+C_{2019}^3+C_{2019}^6+...+C_{2019}^{2019}\right)+z_2\left(C_{2019}^1+C_{2019}^4+...+C_{2019}^{2017}\right) \\ +{z_2}^2\left(C_{2019}^2+C_{2019}^5+...+C_{2019}^{2018}\right) \left(1\right) \end{gathered}$$

Tương tự thay $z_3=-\frac{1}{2}-\frac{\sqrt{3}}{2}i$ vào khai triển (*) ta được:

$$\begin{gathered} {\left(-1-\frac{1}{2}-\frac{\sqrt{3}}{2}i\right)}^{2019}=C_{2019}^0+C_{2019}^1{z}_3+C_{2019}^2.{z_3}^2+...+C_{2019}^{2019}{z_3}^{2019} \\ \iff {\left(\frac{1}{2}-\frac{\sqrt{3}}{2}i\right)}^{2019}=C_{2019}^0+C_{2019}^1{z}_3+C_{2019}^2{z_3}^2+C_{2019}^3+...+C_{2019}^{2019} \\ \iff -1=\left(C_{2019}^0+C_{2019}^3+C_{2019}^6+...+C_{2019}^{2019}\right)+z_3\left(C_{2019}^1+C_{2019}^4+...+C_{2019}^{2017}\right) \\ +{z_3}^2\left(C_{2019}^2+C_{2019}^5+...+C_{2019}^{2018}\right) \left(2\right) \end{gathered}$$

Thay $z=1$ vào khai triển (*) ta được:

$$\begin{gathered} 2^{2019}=C_{2019}^0+C_{2019}^1+C_{2019}^2+...+C_{2019}^{2019} \\ \iff 2^{2019}=\left(C_{2019}^0+C_{2019}^3+C_{2019}^6+...+C_{2019}^{2019}\right)+\left(C_{2019}^1+C_{2019}^4+C_{2019}^7+...+C_{2019}^{2017}\right) \\ +\left(C_{2019}^2+C_{2019}^5+C_{2019}^8+...+C_{2019}^{2018}\right) \left(3\right) \end{gathered}$$

Cộng vế với vế của (1); (2); (3) ta được:

$$\begin{gathered} 2^{2019}-2=3\left(C_{2019}^0+C_{2019}^3+...+C_{2019}^{2019}\right)+\left(1+z_2+z_3\right)\left(C_{2019}^1+C_{2019}^4+...+C_{2019}^{2018}\right) \\ +\left(1+{z_2}^2+{z_3}^2\right)\left(C_{2019}^2+C_{2019}^5+...+C_{2019}^{2017}\right) \end{gathered}$$

Nhận thấy $1+z_2+z_3=1-\frac{1}{2}+\frac{\sqrt{3}}{2}i-\frac{1}{2}-\frac{\sqrt{3}}{2}i=0$ và $1+{z_2}^2+{z_3}^2=1+{\left(-\frac{1}{2}+\frac{\sqrt{3}}{2}i\right)}^2+{\left(-\frac{1}{2}-\frac{\sqrt{3}}{2}i\right)}^2=0$

Nên $2^{2019}-2=3S\iff S=\frac{2^{2019}-2}{3}$.