Đáp án: $1+2^{1009}$
Giải thích các bước giải:
Ta có:
$z=1+(1+i)+(1+i)^2+..+(1+i)^{2018}$
$\to z=1+\dfrac{1-(1+i)^{2019}}{1-(1+i)}$
$\to z=1+\dfrac{1-(1+i)^{2019}}{-i}$
Lại có:
$(1+i)^{2019}=((1+i)^2)^{1009}(1+i)=(2i)^{1009}(1+i)=2^{1009}\cdot i^{1009}(1+i)= 2^{1009}\cdot (i^4)^{252}\cdot i(1+i)= 2^{1009}\cdot 1^{252}\cdot i(1+i)$
$\to (1+i)^{2019}=2^{1009}(i+i^2)$
$\to (1+i)^{2019}=2^{1009}(i-1)$
$\to z=1+\dfrac{1-2^{1009}(i-1)}{-i}$
$\to z=1+\dfrac{1-2^{1009}i+2^{1009}}{-i}$
$\to z=1+\dfrac{i-2^{1009}i^2+2^{1009}i}{-i^2}$
$\to z=1+\dfrac{i+2^{1009}+2^{1009}i}{1}$
$\to z=1+(i+2^{1009}+2^{1009}i)$
$\to z=(1+2^{1009})+i(1+2^{1009})$
$\to$Phần ảo là $1+2^{1009}$
