Answer
\[\overline{z_1+z_2+ ...+ z_n }=\bar{z_1}+\bar{z_2}+...\bar{z_n}\]
Work Step by Step
Let $z_1,z_2, , ..., z_n $ are complex numbers
Claim:- \[\overline{z_1+z_2+ ...+ z_n }=\bar{z_1}+\bar{z_2}+...\bar{z_n}\]
Because $z_1,z_2, , ..., z_n $ are complex numbers
\[z_1=a_1+ib_1\;\;\;\]
\[z_2=a_2+ib_2\]
\[\vdots\]
\[z_n=a_n+ib_n\]
Where $a_1,a_2,...,a_n$ and $b_1,b_2,...,b_n$ are real numbers
\[z_1+z_2+ ...+ z_n =(a_1+ib_1)+(a_2+ib_2)+...+(a_n+ib_n)\]
\[z_1+z_2+ ...+ z_n =(a_1+a_2+...+a_n)+i(b_1+b_2+...+b_n)\]
\[\overline{z_1+z_2+ ...+ z_n }=(a_1+a_2+...+a_n)-i(b_1+b_2+...+b_n)\]
\[z_1+z_2+ ...+ z_n =(a_1-ib_1)+(a_2-ib_2)+...+(a_n-ib_n)\]
\[\overline{z_1+z_2+ ...+ z_n }=\bar{z_1}+\bar{z_2}+...\bar{z_n}\]
Hence, \[\overline{z_1+z_2+ ...+ z_n }=\bar{z_1}+\bar{z_2}+...\bar{z_n}\]
