Without adding, find the sum:
(i) (1+3+5+7+9+11+13)
(ii) (1+3+5+7+9+11+13+15+17+19)
(iii) (1+3+5+7+9+11+13+15+17+19+21+23)

We know that sum of the first n odd natural numbers= $n^2$
Therefore,
(i) $\because$ It ends with 13
and 1+3+5+7+9+11+13 is the sum of first 7 odd numbers
$\therefore$ Its sum= $(7{)}^2=49$
(ii) (1+3+5+7+9+11+13+15+17+19)
Here, n=10
$\therefore$ Sum= $n^2=(10{)}^2=100$
(iii) (1+3+5+7+9+11+13+15+17+19+21+23)
Here, n=12
$\therefore$ Sum= $n^2=(12{)}^2=144$