In the SQL language. Find 1 2 3 ... 100 of the sum. Thank you

To add a note: the following examples have passed my test, so you can use them with confidence.

Oracle.

set serveroutput on;

declare

sums int;

i int;

beginsums:=0;

i:=1;while i<=100

loopsums:=sums+i;

i:=i+1;

end loop;

dbms_'Integers within 100 and :'||sums);end;

sqlserver.

declare @i int;

declare @sums int;

set @i=1;

set @sums=0;

while (@i<=100)

beginset @sums=@sums+@iset @i=@i+1

endselect @sums

Above, I hope it will help you.

Create a stored procedure.

create proc addtotal

asdeclare @num int,@result intbeginset @num=1

set @result=0

while 1=1

beginif @num>100

breakelse

beginset @result=@result+@numset @num=@num+1

endend

print 'The result of the calculation is:'+convert(varchar(10),@result)

endexec addtotal

s=a1+(a1+a2)+(a1+a2+a3)s=b1+b2+b3

a2=a1+2

a3=a2+2

--Get the general formula an=a(n-1)+2b1=a1

b2=2*a1+2

b3=3*a1+6

Then we can get that the general term of bn is bn=n(a1+n-1), since a1=1, the value of bn is equal to the square of n, bn=n2, so.

s=1^2+2^2+3^2+..The formula for the equation n 2 is (I have already returned it to the teacher after referring to the source) n(n+1)(2n+1) 6

So s=n(n+1)(2n+1) 6

So as long as the value of n is determined, it is only an expression to express it in sql, assuming n=3:

select