Question #45c04

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Answer 1 · 2018-01-02T13:34:20

See the proof below

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 $"n"$