MySQL 5.6

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select version();


version()


5.6.45


create table tab( id int, daily int )
insert into tab values
(1,2000),
(2,2450),
(3,3500);
select t.id, t.daily,
       (select sum(daily) from tab where id <= t.id ) as accumulation
  from tab t
 order by t.id;
  


id
daily
accumulation


1
2000
2000

2
2450
4450

3
3500
7950