We must prove the statement #P(n)# that
#bar(z_1+z_2+z_3+.......z_n)=bar(z_1)+bar(z_2)+bar(z_3)+.......bar(z_n)#
#"Proof by Induction"#
#(1)# When #n=1#
#P(1)#, #=>#, #bar(z_1)=bar(z_1)#
The statement is true for #P(1)#
#"Inductve step"#
Suppose that the statement is true for #P(n)#
#bar(z_1+z_2+z_3+.......z_n)=sum_(k=1) ^nbar(z_k)#
Then,
#bar(z_1+z_2+z_3+.......z_n)+bar(z_(n+1))=(sum_(k=1) ^nbar(z_k))+bar(z_(n+1))#
#=sum_(k=1) ^(n+1)bar(z_k)#
#=bar(z_1+z_2+z_3+.......z_(n+1)#
Therefore,
the statement is true for #P(n+1)#
#"Conclusion : "# The statement is true for #"n=1"#, #"n"# and #"(n+1)"#, we conclude that the statement is true for all #"n"#
