adición de fecha en la función sql

La forma más fácil de resolver su pregunta es usar ROLLUP en grupo por consulta (http://technet.microsoft.com/en-us/library/bb522495(v=sql.105).aspx )

Aquí está SQL Fiddle (http://sqlfiddle.com/#!3/3ce09/19 )

Definición de fecha:

create table test (
    Id  INT IDENTITY(1,1),
    date  DateTime,
    balls  int );
insert into test(date, balls) values('11/21/2013',  2);
insert into test(date, balls) values('11/22/2013',  3);
insert into test(date, balls) values('11/23/2013',  4);
insert into test(date, balls) values('11/24/2013',  5);
insert into test(date, balls) values('11/25/2013',  2);
insert into test(date, balls) values('11/26/2013',  5);
insert into test(date, balls) values('11/27/2013',  4);
insert into test(date, balls) values('11/28/2013',  3);
insert into test(date, balls) values('11/29/2013',  9);
insert into test(date, balls) values('11/30/2013',  8);
insert into test(date, balls) values('12/01/2013',  7);
insert into test(date, balls) values('12/02/2013',  4);
insert into test(date, balls) values('12/03/2013',  5);
insert into test(date, balls) values('12/04/2013',  6);
insert into test(date, balls) values('12/05/2013',  2);
insert into test(date, balls) values('12/06/2013',  0);
insert into test(date, balls) values('12/07/2013',  1);

Consulta real:

select 
  label, 
  balls 
  from ( 
     SELECT  
       (DATEDIFF(d , '11/21/2013' , date) / 7) as week, 
       ((DATEDIFF(d , '11/21/2013' , date) / 7) + 1) as week1, 
       LEFT(CONVERT(VARCHAR, (DateAdd(d, (DATEDIFF(d , '11/21/2013' , date) / 7)*7, '11/21/2013')), 120), 10) +    ' - ' +  
       LEFT(CONVERT(VARCHAR, (DateAdd(d, ((DATEDIFF(d , '11/21/2013' , date) / 7) + 1)*7 - 1, '11/21/2013')), 120), 10) as label, 
       SUM(Balls) as balls 
      FROM test 
      GROUP BY rollup((DATEDIFF(d , '11/21/2013' , date) / 7))   
  ) as t 
  where t.label is not null

Resultado:

LABEL                       BALLS
2013-11-21 - 2013-11-27     25
2013-11-28 - 2013-12-04     42
2013-12-05 - 2013-12-11     3