(1) 1+ 5+ 11+ 19+...
1st diff: 4 6 8
2nd diff: 2 (constant)
S[n]=a+bn+cn^2+dn^3 where a, b, and d are constants.
S=1 when n=1: 1=a+b+c+d (A)
S=6 when n=2: 6=a+2b+4c+8d (B)
S=17 when n=3: 17=a+3b+9c+27d (C)
S=36 when n=4: 36=a+4b+16c+64d (D)
(B)-(A): 5=b+3c+7d (E)
(C)-(B): 11=b+5c+19d (F)
(D)-(C): 19=b+7c+37d (G)
(F)-(E): 6=2c+12d, 3=c+6d (H)
(G)-(F): 8=2c+18d, 4=c+9d (I)
(I)-(H): 1=3d so d=1/3, 4=c+3 so c=1, 5=b+3+7/3, b=-1/3, a=1-(b+c+d)=1-(-1/3+1+1/3)=0
So S[n]=-n/3+n^2+n^3/3=(n/3)(-1+3n+n^2)=(n/3)(n^2+3n-1).
(2) 2+6+12+20+...=2(1+3+6+10+...)
1st diff: 4 6 8 10
2nd diff: 2 (constant)
S[n]=A+Bn+Cn^2+Dn^3
Eliminate A because S0=0
S1=2=B+C+D (a)
S2=8=2B+4C+8D (b)
S3=20=3B+9C+27D (c)
(b)-2(a): 4=2C+6D, 2=C+3D so C=2-3D
(c)-3(a): 14=6C+24D, 7=3C+12D=6-9D+12D=6+3D, 3D=1 so D=1/3, 7=3C+4, C=1 and B=2-4/3=2/3.
S[n]=2n/3+n^2+n^3/3=(n/3)(2+3n+n^2)=(n/3)(n+2)(n+1)