# Question 45c04

Jan 2, 2018

See the proof below

#### Explanation:

We must prove the statement $P \left(n\right)$ that

$\overline{{z}_{1} + {z}_{2} + {z}_{3} + \ldots \ldots . {z}_{n}} = \overline{{z}_{1}} + \overline{{z}_{2}} + \overline{{z}_{3}} + \ldots \ldots . \overline{{z}_{n}}$

$\text{Proof by Induction}$

$\left(1\right)$ When $n = 1$

$P \left(1\right)$, $\implies$, $\overline{{z}_{1}} = \overline{{z}_{1}}$

The statement is true for $P \left(1\right)$

$\text{Inductve step}$

Suppose that the statement is true for $P \left(n\right)$

$\overline{{z}_{1} + {z}_{2} + {z}_{3} + \ldots \ldots . {z}_{n}} = {\sum}_{k = 1}^{n} \overline{{z}_{k}}$

Then,

$\overline{{z}_{1} + {z}_{2} + {z}_{3} + \ldots \ldots . {z}_{n}} + \overline{{z}_{n + 1}} = \left({\sum}_{k = 1}^{n} \overline{{z}_{k}}\right) + \overline{{z}_{n + 1}}$

$= {\sum}_{k = 1}^{n + 1} \overline{{z}_{k}}$

=bar(z_1+z_2+z_3+.......z_(n+1)#

Therefore,

the statement is true for $P \left(n + 1\right)$

$\text{Conclusion : }$ The statement is true for $\text{n=1}$, $\text{n}$ and $\text{(n+1)}$, we conclude that the statement is true for all $\text{n}$