text: 1+3+5+7+9+...........+(2n-1)  where   n=1,2,3,4,....\n\nIn AP first term a=1 and diff d=2\n\nAs we know in AP  (d-a)+(2d-a)+.........+(nd-a) if d>a and n=1,2,3,..\nsum= n(n+1)d/2 -na    where  n=1,2,3..\n substituting a=1 and d=2 ( for odd numbers)\nsum=n^2+n-n\nsum=n^2.