Chapter 5.2, Problem 5ES

Fill in the missing pieces in the following proof that $1+3+5+\cdot \cdot \cdot \left(2n-1\right)=n^2$ for every integer $n\ge 1$ .
Proof: Let the property P(n) be the equation
$1+3+5+\text{···}+\left(2n-1\right)=n^2.\leftarrow P(n)$
Show that P(1) is true: To establish P(l), we must show that when 1 is substituted in place of n, the left-hand side equals the right-hand side. But when $n=1$ , the left-hand side is the sum of all the odd integers from 1 to 2 • 1 — 1, which is the sum of the odd integers from 1 to 1 and is just 1. The right-hand side is $\underset{\_}{(a)}$ which also equals 1. So P(1) is true.
Show that for every integer $k\ge 1,$ P(k) is true then $P\left(k+1\right)$ is true: Let k be any integer with $k\ge 1$ .
[Suppose P(k) is true. That is:] Suppose
$1+3+5+\cdot \cdot \cdot +\left(2k-1\right)=\overline{\left(b\right)}.\leftarrow P\left(k\right)$
[This is the inductive hypothesis.]   [We must show that $P(k+1)$ is true. Thar is:] We must show that $\overline{\left(c\right)}=\overline{\left(d\right)}\leftarrow P\left(k+1\right)$
Now the left-hand side of $P(k+1)$ is

which is the right-hand side $P\left(k+1\right)$ [as was to be shown].
[Since we have proved the basis step and the inductive step, we conclude that the given statement is true.]
Note: This proof was annotated to help make its logical flow more obvious. In standard mathematical writing, such annotation is omitted.
Prove each statement in 6—9 using mathematical induction. Do not derive them from Theorem 5.2.1 or Theorem 5.2.2.