Write out a bunch of the sums:

1*2 = 2
1*2 + 2*3 = 2 + 6 = 8
1*2 + 2*3 + 3*4 = 8 + 12 = 20
1*2 + 2*3 + 3*4 + 4*5 = 20 + 20 = 40
1*2 + 2*3 + 3*4 + 4*5 + 5*6 = 40 + 30 = 70
1*2 + 2*3 + 3*4 + 4*5 + 5*6 + 6*7 = 70 + 42 = 112

Make a difference table:

List those partial sum in a column.  Subtract each number
from the number just below it, and write that number to
the right of that number.  To get the 6 at the top of the 2nd
column we subtract the 2 from the 8 in the 1st column and
wrote it just right of the 2.  To get the 42 at the bottome 
of the 2nd column we subtracted the 70 from the 112 below it  

  2  6  6  2
  8 12  8  2
 20 20 10  2 
 40 30 12
 70 42
112

We continue to make the chart until we get a column with
all the same numbers in it.  Since it took three columns 
to get a column all the same number we know to assume that 
the formula for the nth term is a 3rd degree polynomial 
in n:

S(n) = An³ + Bn² + Cn² + D

There are four unknown constants so we will substitute the
first 4 terms of the sequence:

S(1) = A(1)³ + B(1)² + C(1)² + D = A + B + C + D = 2
S(2) = A(2)³ + B(2)² + C(2)² + D = 8A + 4B + 2C + D = 8
S(3) = A(3)³ + B(3)² + C(3)² + D = 27A + 9B + 3C + D = 20
S(4) = A(4)³ + B(4)² + C(4)² + D = 64A + 16B + 4C + D = 40

  A +   B +  C + D =  2
 8A +  4B + 2C + D =  8
27A +  9B + 3C + D = 20
64A + 16B + 4C + D = 40

Subtracting the 1st equation from the 2nd:

7A + 3B + C = 6

Subtracting the 2nd equation from the 3rd:

19A + 5B + C = 12

Subtracting the 3rd equation from the 4th:

37A + 7B + C = 20

So now we have the system

 7A + 3B + C =  6
19A + 5B + C = 12
37A + 7B + C = 20

Subtracting the 1st equation from the 2nd:

12A + 2B = 6, dividing thru by 2, 6A + B = 3

Subtracting the 2nd equation from the 3rd:

18A + 2B = 8, dividing thru by 2, 9A + B = 4


So now we have the system

6A + B = 3
9A + B = 4

Subtracting the 1st equation from the 2nd:

3A = 1
 A = 

Substitute in 

6A + B = 3
6( + B = 3
2 + B = 3
    B = 1

Substitute in 
  
 7A + 3B + C =  6
7( + 3 + C = 6
Multiply thru by 3
7 + 9 + 3C = 18
16 + 3C = 18
     3C = 2
      C = 

Substitute in 

A +  B + C + D =  2
 + 1 +  + D = 2
Multiply thru by 3
1 + 3 + 2 + 3D = 6
6 + 3D = 6
    3D = 0
     D = 0

So the formula is

S(n) = An³ + Bn² + Cn² + D

S(n) = n³ + 1n² + n + 0

Multiply thru by 3

3S(n) = n³ + 3n² + 2n

3S(n) = n(n² + 3n + 2)

3S(n) = n(n+1)(n+2)

 S(n) = n(n+1)(n+2)

Edwin