Cho khai triển (2018x^2 + x + 6055)^2018 = a0+ a1+

* Hướng dẫn giải

Chọn đáp án A

Thay $x=i\sqrt{3}$, ta có:

${\left(1+i\sqrt{3}\right)}^{2018}=a_0+a_{1}i\sqrt{3}+a_2{\left(i\sqrt{3}\right)}^2+a_3{\left(i\sqrt{3}\right)}^3+...+a_{4036}{\left(i\sqrt{3}\right)}^{4036}$

$\iff {\left[{\left(1+i\sqrt{3}\right)}^3\right]}^{672}{\left(1+i\sqrt{3}\right)}^2$

$=\left(a_0-3a_2+3^2{a}_4-...+3^{2018}{a}_{4036}\right)+i\sqrt{3}\left(a_1-3a_3+3^2{a}_5-...-3^{2017}{a}_{4035}\right)$

$\iff 8^{672}\left(-2+2i\sqrt{3}\right)$

$=\left(a_0-3a_2+3^2{a}_4-...+3^{2018}{a}_{4036}\right)+i\sqrt{3}\left(a_1-3a_3+3^2{a}_5-...-3^{2017}{a}_{4025}\right)$

$\Rightarrow a_0-3a_2+3^2{a}_4-...+3^{2018}{a}_{4036}=-2.8^{672}=-2^{2017}$