Why is the sum of n terms of an AP equal to a relation that isconstant for all APs

The sum of n terms of an AP  Sn= {2a1+(n-1)d}n/2 is a
relation, connecting the  sum  Sn  with a1, the starting term, n the number of terms and
d the common difference.

The sum Sn is dependent on n the
variable , a1 , and d in the AP like:

Sn = {(2a1+(n-1)d}n/2
= (d/2)n^2 + (a1-d/2)n  is a  second degree  in n.

Here a1
is the first term , ar = a1+(r-1)d th term of the AP

The
relation is unique or constant.

Now let us
test:

Let  n= 1.

LHS S1 =1.
RHS = {2a1 +a1(1-1)d}*1/2 = (2a1+0*d}1/2 =a1.

S2 =
{2a1+a1(2-1)d}/2 = a1+a1+d = a1+(a1+d) = a1+a2 holds
good.

S3 = (2a1 +a1(3-1)d)3/2 = 3a1+2d = a1+(a1+d)+(a1+2d)
= a1+a2+a3 holds good.

S4 = {2a1+a1(4-1)d}4/2 = 4a1+3d =
a1+(a1+d)+(a1+2d)+(a1+3d) = a1+a2+a3+a4.

So  Sn =
( d/20n^2+(a1-d/2)n a second degree function (or relation) in n holds true for more than
2 values. So the relation is unique or an identity. Therefore the relation is
constant.