Answer
The sum of the first n positive odd integers
1 + 3 + 5 + . . . . . . . . . +(2n - 1) = $n^{2}$
Proof-
$1^{st}$ term $a_{1}$ = 1
Common difference = 2
Last term $a_{L}$ = (2n - 1)
The sum $S_{n}$ = $\frac{n}{2}$(first term + last term)
= $\frac{n}{2}$($a_{1}$ + $a_{L}$)
= $\frac{n}{2}$(1 + 2n - 1)
= $\frac{n}{2}$(2n)
=n$\times$n
=$ n^{2}$
Work Step by Step
The sum of the first n positive odd integers
1 + 3 + 5 + . . . . . . . . . +(2n - 1) = $n^{2}$
Proof-
$1^{st}$ term $a_{1}$ = 1
Common difference = 2
Last term $a_{L}$ = (2n - 1)
The sum $S_{n}$ = $\frac{n}{2}$(first term + last term)
= $\frac{n}{2}$($a_{1}$ + $a_{L}$)
= $\frac{n}{2}$(1 + 2n - 1)
= $\frac{n}{2}$(2n)
=n$\times$n
=$ n^{2}$
